CALCULUS 1: LIMITS, AVERAGE GRADIENT AND FIRST PRINCIPLES DERIVATIVES

Size: px
Start display at page:

Download "CALCULUS 1: LIMITS, AVERAGE GRADIENT AND FIRST PRINCIPLES DERIVATIVES"

Transcription

1 6 LESSON CALCULUS 1: LIMITS, AVERAGE GRADIENT AND FIRST PRINCIPLES DERIVATIVES Learning Outcome : Functions and Algebra Assessment Standard 1..7 (a) In this section: The limit concept and solving for limits Function notation (brief revision section) Finding the rate of change of a function between two points or average gradient Calculating the derivative of a function from first principles Limit Concept Background The concept of finding a limit looks at what happens to a function value (-value) of a curve, as we get close to a specific -value on the curve. One wa to think of it is: if ou start walking in a certain direction and find ourself heading towards a big boulder, this boulder will be our limit. Eample Eample 1 Let s start b looking at the function ƒ() = shown below: f() 6 Now let s consider the lim (we sa: the limit of ƒ() as tends to zero). So, 0 we want to know what the -value is equal to when approaching = 0. Page 88

2 We sa: lim ƒ() = lim = 3(0) + 6 Substitute in = 0, and remove lim 0 = 6 Eample In the case of ƒ() = _ 16 (shown below) 4 we can use a limit to find what value ƒ() tends to as tends to four. 8 f() Eample But if we substitute = 4 into the limit, we will get lim ƒ() = _ 0, but we know that this 4 0 is wrong. So how can we solve this algebraicall? Well, let s see if we can simplif ƒ() at all: ƒ() = _ The numerator is a difference of two squares, so we can factorise ( 4)( + 4) = The ( 4) in the numerator and denominator will cancel 4 = + 4 Now let s solve for the limit of ƒ(): lim ƒ() = lim 4 4 = lim 4 _ Now we can substitute in for = 4 = 8 this means that ƒ() tends to 8 but does not equal 8 at the value of 4 Note that if substituting for the limit produces a zero denominator, factorise and cancel first! Activit 1 Activit 1) lim 5 + ) lim ) lim 1_ + + 1_ Page 1 Lesson 1 Algebra Page 89

3 Revision of Function Notation Eample Eample Eample 1 If ƒ() = 3 +, find the value of: 1. ƒ() ƒ() tells us that in the formula ƒ(), we must make the substitution = ƒ() = () 3() + = = 0 this tells us that on the curve ƒ(), at the point =, ƒ() = 0. In actual fact we have found the co-ordinates (; 0) of a point on the curve ƒ().. ƒ( 1) ƒ( 1) tells us to make the substitution = 1 in ƒ() ƒ( 1) = ( 1) 3( 1) + = = 6 3. ƒ( + ) ƒ( + ) we substitute = + into ƒ() ƒ( + ) = ( + ) 3( + ) + multipl out = simplif = + 4. ƒ( + h) in this case we must substitute = + h into f(). ƒ( + h) = ( + h) 3( + h) + multipl out = + h + h 3 3h + this cannot be simplified further and is therefore our final answer Eample If ƒ() = ƒ( + h) ƒ() +, determine : h ƒ( + h) ƒ() h = ( + h) + ( + h) ( + ) Substitute = + h for ƒ( + h) h = + h + h + + h Remember to distribute the negative sign h h( + h + 1) = Factorise and take a common factor of h h = + h + 1 This cannot be simplified further and is our final answer Page 90

4 Average Gradient Background When dealing with straight line graphs, it is eas to determine the slope of the same graph b calculating the change in -value between two points and the change in -value between the same two points. Then we substitute these into change in the formula for gradient =. However, when dealing with curves the change in gradient changes at ever point on the curve, therefore we need to work with the average gradient. The average gradient between two points is the gradient of a straight line drawn between the two points. Eample Given the equation ƒ() = 6, find the average gradient between = 1 and = 4. First we need to find the value of ƒ() at = 1. This is the -value of ƒ() at = 1. ƒ(1) = (1) 1 6 (making the substitution = 1) = 7 = 5 Net we must find the value of ƒ() at = 4 ƒ(4) = (4) 4 6 = (16) 10 = Now we can substitute these values into the average gradient formula: _ B A B = ( 5) A 4 1 = _ 7 3 = 9 Eample The Derivative (First Principles) Background We will now investigate the average gradient between points. B calculation we can show that this is equivalent to: ƒ( + h) ƒ() h f() f() ( ; f()) f( + h) ( + h ; f( + h)) + h 009 Page 1 Lesson 1 Algebra Page 91

5 Now think about it: as h, the distance between our -values, gets smaller our points get closer together. Lets look at an eample of this: let = for the graph below. ( h = 3, since 5 = 3) for h = 3: f() f() f( + h) h = 3 h 5 g() for h = 1: f() f() f( + h) h = 1 h 3 g() for h = 0.5 f() f() f( + h) h.5 g() h = 0.5 Page 9 In each of the graphs above, g() is a straight line with the same gradient as the average gradient of the curve between the points and + h. Can ou see that as h decreases the two points on the ais get closer together? Can ou also see that as the two poins on the -ais get closer together, the average gradient changes? So as the two points get closer together (as h becomes 0) the average gradient between the two points becomes the actual gradient at the point. However, h

6 cannot equal 0 as that would give us a 0 denominator, so we must use limits. The actual gradient at a point is called the derivative. To find the limit of the gradient as h tends to zero, we use the formula: lim ƒ( + h) f(). We call this the derivative of the function and use the h 0 h notation ƒ () This method is called finding the derivative from first principles. Eample Find the derivative of the function ƒ() = 3 from first principles. Solution: ƒ( + h) = 3( + h) = 3( + h + h ) = 3 + 6h + 3h Start b finding f( + h), which is the first part of the formula Now substitute the epressions for ƒ() and ƒ( + h) into the first principles formula: Eample Solution ƒ( + h) ƒ() ƒ () = lim h h 0 (3 + 6h + 3h ) (3 ) h 0 h 3 + 6h + 3h 3 h 0 h = lim = lim 6h + 3h = lim h h 0 ƒ () = lim h(6 + 3h) h 0 h Dont forget to change the signs of the term in the second bracket This can be simplified further If we were to evaluate this now, we would get ƒ () = _ 0. But we 0 have a common factor of h in the numerator, so let s factorise. The h in the numerator will cancel out the h in the denominator = lim (6 + 3h) h 0 = 6 +3(0) Substitute in for h = 0, and remove lim h 0 = 6 This is the derivative of ƒ() = 3 Activit Activit 1. For the curve ƒ() = - + 9, find: 1.1 the average gradient between = 0 and = 6 1. the derivative, from first principles, at = 3. From first principle, find ƒ () for the following functions:.1 ƒ() = Page 1 Lesson 1 Algebra Page 93

7 . ƒ() = Calculus : General Rules for Differentiation, and Tangents Learning Outcome : Functions and Algebra Assessment Standard 1..7 (a) In this section we will: determine the general rules for differentiation determine the gradient and equation of a tangent to a function at a desired point General Rules for Differentiation Note: there are man different was of writing ƒ (), we can use: D [ƒ()] we sa: the derivative of ƒ() with respect to _ d d we sa: the derivative of with respect to or f we sa: the derivative of or f() Derivative Laws I am sure ou will agree that finding the derivative from first principles is a rather time-consuming process and that a more time-efficient method is needed. Luckil for ou, there is one! A set of rules have been created in order to speed up the process, and these are: 1. If ƒ() =k n where k is a constant, then ƒ () = k n n 1 If ƒ() = 6 3, then ƒ() = 6 3 = 18. If ƒ() = k, where k is a constant, then ƒ () = 0 If the ƒ() is equal to a constant, then the derivative of the function is zero. If ƒ() = 3, then ƒ () = 0 3. If ƒ() = g() + h(), then ƒ () = g () + h () The derivative of a sum is the sum of the derivatives Ifƒ() = , then ƒ () = Similarl, if ƒ() = g() h(), then ƒ () = g () h () The derivative of a difference, is the difference of the derivatives Ifƒ() = 6 3 6, then ƒ () = Page 94

8 Remember: 1_ = If ou have in the denominator, take it to the numerator and change the sign of the eponent 3 _ = 1_ 3 If ou have a surd, change it to power form. 4 + = _ If ou have man terms over a common denominator, with a single term in the denominator, separate the terms out into several fractions Eamples: Find the derivatives of the following functions 1.1 ƒ() = f ()= ƒ() = _ + 4 f () = _ 1.3 ƒ() = 3 _ = 3 - Take the to the numerator b changing the sign of the eponent ƒ () = 3( 3 ) = 6 3 = _ ƒ() = _ = _ ƒ () = = 4 + _ 4 Activit 3 Eample Activit Find the derivatives of the following functions: _ 1. ƒ() = + 1 _. = 3 ( 3) 3. = Page 1 Lesson 1 Algebra Page 95

9 Derivatives are useful in man aspects of life, one eample is: A If a mountaineer was deciding which mountain he is capable of climbing, he would want to know two things about the mountain: how high it is and how steep it is. If we consider the mountain as a parabola, it would be represented as seen above. Then the height of the mountain at point A could be determined b substitution of the -value of A into ƒ(). The steepness of the slope of the mountain could be found b taking the derivative of the parabola and evaluating it at point A. This is eqiuvalent to finding the gradient of the tangent at A. Finding the Equation of the Tangent Background The tangent to a point on the curve is a straight line that touches the curve at that single point and has the eact same gradient as the curve at that point. tangent f() To find the gradient of the tangent, we need to find the gradient of the curve at that eact point. To do this, we find the derivative of the function at that point (b substituting the -value into the derivative) and this equals the gradient. Remember a tangent is a straight line and so has equation = m + c. We alread have m (the gradient) as shown above and to get c, we simpl substitute the coordinates of the point of contact between tangent and curve into the equation = m + c. So that s how we find the equation of a tangent to a curve using the derivative. Eample Eample 1 If we look at the function ƒ() = and want to find the gradient of the tangent at the point = 4, first we need to find the derivative of ƒ(): ƒ () = + This is the general formula for the gradients of all tangents to the curve ƒ(), called the gradient function, but we want to know the specific value of the gradient at = 4, so we must substitute = 4 into the derivative: Page 96

10 m Tangent = (4) + = 10 Therefore, the gradient of the tangent toƒ() at the point = 4 is 10. If we want to find the equation of the tangent, we need a point on the curve. Since we onl know the -value in this case, we need to find the -value at = 4, so we use the original curve to get this -value ƒ (4) = (4) + (4) + 1 making the substitution = 4 into ƒ() = = 5 so we have the point on ƒ(): (4, 5) We know that the general formula for a straight line graph is = m + c. Now that we have found a point on the graph and we know the gradient, we can solve for c: 5 = 10(4) + c substituting in = 5, = 4 and m = 10 5 = 40 + c 5 40 = c c = 15 Therefore, the equation for the tangent to ƒ() at = 4 is = Eample Consider ƒ() = Sketch: ƒ() 1. Determine: ƒ () 1.3 Determine: ƒ (1) 1.4 What does this number represent? 1.5 Hence determine the equation of the tangent to ƒ() at = Determine the equation of the tangent at =. Eample Solution 1.1 Sketch: ƒ() First find intercepts with the aes: Let = 0 to find the -intercept, so = 8 and the -intercept is at (0, 8) To find the -intercepts let = 0, so 0 = ( + )( + 4) and the -intercepts are at ( ; 0) and ( 4 ; 0) Ais of smmetr: = _ b a = _ ( ) ( 1) = _ = 1 For the turning point we must find the -value at the ais of smmetr: = ( 1) ( 1) + 8 = = Solution Page 1 Lesson 1 Algebra Page 97

11 Turning Point ( 1; 9) Now we can sketch ƒ(): ( 1; 9) (0 ; 8) ( 4 ; 0) ( ; 0) 1. Determine: ƒ () ƒ () = 1.3 Determine: ƒ (1) ƒ (1) = (1) substitute = 1 into ƒ () = What does this number represent? The gradient of the tangent at = Hence determine the equation of the tangent to ƒ() at = 1 We have alread found the gradient of the tangent at = 1. Now we need to find a point on ƒ() and solve for the c from = m + c. We need to find the -value of the curve at = 1, so we need to find ƒ(1): ƒ(1) = (1) (1) + 8 = = 5 (1; 5) is the point on ƒ() at which we are tring to find the tangent Now we can use this to solve for c: 5 = (-4)(1) + c substitute in = 5, m = -4 and = 1 5 = -4 + c c = c = 9 = is the tangent to ƒ() at = 1 Page 98

12 1.6 Determine the equation of the tangent at = First we need to find ƒ ( ) ƒ ( ) = ( ) = 4 = This is the gradient of the tangent of ƒ() at = Now we need to find the corresponding -value at this point ƒ( ) = ( ) ( ) + 8 = = 8 ( ; 8) is the point on ƒ() at which we are tring to find the tangent We have everthing we need to solve for c in the equation (substitute the point ( ;8)) = m + c: = + c 8 = 4 + c c = 1 = + 1 is the tangent to ƒ() at = Activit 4 Activit 1. Find the equation of the tangent to the curve ƒ() = ( 1) at the point where the graph meets the -ais. Find the equation of the tangent to the curve ƒ() = + + at the point (1 ; ) 3. The line = is a tangent to the curve of ƒ() = Determine the point of contact. 009 Page 1 Lesson 1 Algebra Page 99

13 4. Find the equation of the tangent to the curve of ƒ() = _ 3 at the point = 1 Calculus 3: Sketching and Analsing Graphs Using Calculus Learning Outcome : Functions and Algebra Assessment Standard 1..7 (a) In this section: Sketching a cubic graph Finding the equation of a cubic graph How to work with points of Inflections Application of maima and minima in real-life scenarios Calculating the rate of change in real-life scenarios Cubic Graphs Background The cubic graph has a general formula: ƒ() = a 3 + b + c + d. Let s first look at some new concepts and terminolog that must be learnt: Stationar point: a point on the function where the gradient is zero. There are three kinds of stationar points: 1. Local maimum: f () = 0 local maimum positive gradient negative gradient. Local Minimum: negative gradient positive gradient local minimum f () = 0 Page 100

14 3. Point of Inflection: f BE CAREFUL: Not all points of inflection are stationar points! The gradient of the curve MUST BE ZERO at a stationar point. As with curve sketching in the past: For -intercept, put = 0 then solve for, i.e. = d. Intercept at (0; d) For -intercepts, put = 0, i.e. solve a 3 + b + c + d = 0. There could be one, two or three real roots. For stationar points, put ƒ () = 0 and solve for, i.e. 3a + b + c = 0, then use these values to find the -value of the curve at these points. Find the points of inflection b setting the second derivative equal to zero. Points of inflection, f () = 0 There could be: TWO stationar points, in which case the possible graphs for a>0 are: A B C 3 real roots turning points A B real roots turning points A 1 real root turning points And for a < 0: A B C 3 real roots turning points A real roots turning points B 1 real root turning points ONE stationar point, in which case the possible graphs are: A 009 Page 1 Lesson 1 Algebra Page 101

15 NO stationar points: in this case ou ma have a point of inflection for which the gradient is not zero. Eample Eample 1 (Curve Sketching) ƒ() = Sketch ƒ(), clearl showing all intercepts with the aes and stationar points. Stationar Points: A good wa to start is to find the stationar points. In order to do this, we must find ƒ () = 0 since at the minimums and maimums, the gradient of the curve is 0. ƒ () = = 0 factorise (3 + 1)( 1) = 0 = _ 1 3 OR = 1 Now we need to find the -values of the curve at the stationar points, so we must find ƒ ( _ 1 3) and ƒ(1). Substitute = _ 1 into ƒ() Substitute = 1 into ƒ() 3 f( _ 1 3 ) = ( _ 1 3) 3 ( _ 1 3) ( _ 1 3) + 10 ƒ(1) = = _ 1 7 _ _ = 9 = (1; 9) 7 = _ 75 7 = 10 _ , ( _ 1 3 ; 10 5_ 7) Page 10

16 These are the two points on ƒ() at which we have stationar points. From the stationar points, we can now make a rough sketch of the graph: Points of Inflection: to find if there are an points of inflection, we need to solve for ƒ () = 0 ƒ () = 6-6 = 0 6 = = _ 1 3 \ there is a point of inflection at = _ 1 3. We need to find f( _ 1 3 ) f( _ 1 3 ) = ( _ 1 3 )3 (_ 1 3 ) (_ 1 3 ) + 10 = _ 1 7 _ 1 9 _ = = _ 59 7 = (_ 1 3 ; _ 59 7 ) -intercepts: Net we must find the -intercepts b making ƒ() = 0 Cubic functions are more difficult to factorize than quadratic functions. Using the factor theorem is often the easiest wa. REMEMBER: when using the factor theorem, if we find ƒ(a) = 0 for some value, a, then ( a) is a factor of ƒ(). Use trial and error to get our first factor. Tr ƒ( 1) ƒ(1), ƒ(-), and ƒ() until ou get one equal to zero ƒ(1); ƒ( 1) 0; ƒ() 0 but ƒ( ) = = 0 + is a factor. We usuall get the other factor(s) b doing long division, snthetic division or the inspection method. B looking at the rough sketch above can ou see wh there is onl 1 -intercept -intercept: to find the -intercept, let = 0 = (0) 3 (0) = Page 1 Lesson 1 Algebra Page 103

17 Now we can sketch the graph b joining all the points: note that if coefficient of 3 in the original equation is positive, the graph has shape and if the coefficient of 3 is negative,. In this case we have 1 3 so the shape is as follows. ( _ 1 3 ;_ 75 7 ) 10 ( _ 1 3 ;_ 59 7 ) f() (1; 9) _1 _ Eample Eample (Finding the Equation of a Curve) Find the equation of ƒ(): (0 ; 16) f ( ; 0) ( ; 0) Use = a( 1 ) ( ) ( 3 ) Where 1, and 3 are -intercepts This is another form of the general formula of a cubic graph. This form is normall used when finding the equation of a cubic graph as opposed to sketching the graph. The -intercepts are = and =, but at = the graph does not cut the aes, so here we have two -intercepts that are the same. Making this substitution in the general form, we have: = a( + ) ( ) ( ) We are also given the point (0; 16) which is the -intercept. We can substitute this into out general form to find the value of a: 16 = a(0 + )(0 ) (0 ) 16 = a()( )( ) 16 = 8a a = = ( + )( ) Substitute a = and multipl out. = ( + 4)( 4 + 4) = Page 104

18 = Eample 3 (Point of Inflection). Find the points of inflection and stationar points in = If we were to look at the function = and tr to solve for the stationar points of this function we would get: _ d d = = 0 (solve for the stationar points) This cannot be factorised easil, so we use the quadratic formula: = b ± b 4ac a = ± ( ) 4(3)(1) (3) _ = ± _ Eample Now we have a problem, there _ is no solution for as we have 8 in the quadratic equation. Therefore we can sa that there are no stationar points on this curve. However we can find the point of inflection b setting the second point of inflection ( 1 ; ) 3 f() = + + derivative equal to zero. Activit 5 Activit 1. The figure represents the graphs of the functions f() = 4 + and g() = g() F B D 0 A f() C E Determine: 1.1 The lengths of DA, OB, and CF. 009 Page 1 Lesson 1 Algebra Page 105

19 1. The coordinates of the turning point E. 1.3 The value(s) of for which the gradients of ƒ() and g() are equal.. ƒ() = Use a calculator to find the coordinates of the stationar points of ƒ(), correct to one decimal digit.. Use the remainder theorem to find the roots of ƒ() = 0.3 Sketch the curve of ƒ(), showing all the important points clearl. 0.4 Determine the gradient of ƒ() at each of the -intercepts. 3. Find an equation for ƒ() using the sketch: f Page 106

20 4. Calculate all the stationar points and points of Inflection in the curve = Maima and Minima in Calculus Real Life Application There are man situations in life where we need to find the maimum or minimum of some quantit. Often businesses need to find was to minimise costs and maimise productivit. This is where maima and minima come in ver useful. First, let s note some basics that will be useful: often we will get asked to maimise or minimise a certain area, volume, or cost function etc. r Rectangle Circle Perimeter = + Perimeter (circumference) = πr Area = Area = πr h h r Bo Clinder Volume = h Volume = πr h Surface Area = + h + h...(closed) Surface Area = πrh + πr... (closed) Surface Area = + h + h...(open) Surface Area = πrh + πr (open) Background In order to find the maimum or minimum of a function, the derivative of the function must be found and made to equal zero. This is similar to the cubic graphs section, where if we wanted to find the local maimum or minimum we did the same thing. The difference here is that instead of finding a local maimum or minimum of a cubic graph, we ma be finding the maimum or minimum volume, area, cost, profit etc. 009 Page 1 Lesson 1 Algebra Page 107

21 Eample Eample Eample 1 A factor has emploees and makes a profit of P rand per week. The relationship between the profit and number of emploees is epressed in the formula P() = , where is the number of emploees Calculate: 1. The number of emploees needed for the factor to make a maimum profit. First we need to find the derivative of P() P () = Net, to find the maimum profit, we need to make P () = 0 and then solve for (the number of emploees needed to achieve this profit). P () = 0 0 = = 600 = 100 = ±10 Remember that (10) = ( 10), so we must include both the positive and negative answers But it is not possible to have a negative number of people! = 10. The maimum profit Now, all we have to do is substitute = 10 back into the formula for P() P(10) = (10) (10) P(10) = (1000) P(10) = P(10) = 5000 Ma Profit is R5000 Eample You have bought 16m of fencing and would like to use it to enclose a field. If length = h and width =, find h in terms of. What is the maimum area possible for the field? h Perimeter=( length) + ( breadth) so we can use the 16 m to get a first equation. 16 = h + 8 = h + Page 108

22 h = 8 Now that we have epressed h in terms of, we can get an equation for the area that does not contain h. When we form the equation that must be differentiated we must alwas epress it in terms of one unknown (usuall ). Area = length breadth (this is knowledge ou have from grade 9 and 10) = h but we know h = 8 = (8 ) = 8 For maimum area, we need to find the derivative of the area, then make darea _ = 0, and solve for : d darea _ = 8 Now, make this = 0 d 0 = 8 = 8 = 4 Therefore, for the maimum area the width of the field must be 4m. Now substitute this into the equation for the length : h = 8 h= 4 For the maimum possible area of the field, the field must be a square with both a length and a width of 4 m, ie the area = 16 m. Activit 6 Activit 1. The sum of one number and three times another number is 1. Find the two numbers if the sum of their squares is a minimum, and give this minimum value.. A rectangular bo without a lid is made of card board of negligible thickness. The sides of the base are cm b 3 cm, and the height is h cm..1 If the total eternal surface area is 00 cm, show that h = _ 0 _ 3 5. Find the dimensions of the bo for which the volume is a maimum 3. ABCD is a square of side 30 cm. PQRS is a rectangle drawn within the square so that BP = BQ = DS = DR = cm. A P B 3.1 Prove that the area (A) of PQRS is given b A = 60 cm Q 3. What is the maimum possible area of PQRS? S D R C 009 Page 1 Lesson 1 Algebra Page 109

23 Rates of Change Real Life Application In an applications that rel on the speed of cars, planes, or even the speed a ball travels through the air, rates of change can be used to calculate the velocit (rate of change of displacement with time) and acceleration (rate of change of velocit with time) of a moving object. So net time ou re watching the F1, motorbike racing, or even our favourite sportsman or woman in action, ou re watching calculus in motion. Velocit/speed = the derivative of displacement/distance with respect to time Acceleration = the derivative of velocit/speed with respect to time Eample Solution Eample Eample 1 A stone is dropped from the top of a tower. The distance, S metres, which it has fallen after t seconds is given b S = 5t. Calculate the average speed of the stone between t = and t = 4 seconds. Solution: As the question asked for average speed, we need to find the average gradient of the curve defined b S = 5t. So first we need to find the value of S at the two points t = and t = 4: S = 5() S = 5(4) = 5(4) = 5(16) = 0 = 80 S(4) S() Average speed = m/s = _ 4 m/s = _ 60 = 30m/s Eample An object is projected verticall upwards from the ground and its movement is described b S(t) = 11t 16t, where S = distance from the starting point in metres and t = time in seconds. 1. Determine its initial velocit The initial velocit of the object is its velocit at t = 0. This is also the rate of change of distance at t = 0. Therefore we need to find the formula for the derivative of S(t) and evaluate it at t = 0: v(t) = ds _ = 11 3t. Remember velocit is the derivative of distance/ dt displacement. v(0) = 11 3(0) substitute = 0 =11 m/s this is the initial velocit Page 110

24 . Determine its velocit after 3 seconds Now we substitute t = 3 into the formula we have found for the velocit v(3) = 11 3(3) = = 16 m/s this is the velocit after 3 seconds 3. Find the maimum height the object reaches This will be the maimum displacement of the object which, from maima and minima, we can find b putting s (t) = 0 and solving for t: S (t) = 11 3t 11 3t = 0 put equal to zero to find t at maimums 11 = 3t t = _ 11 3 = 7_ = 3.5 seconds S (_ 7 ) = 11 (_ 7 ) 16 (_ 7 ) substitute t = 3.5 into S(t) to find maimum height = = 196 It reaches a height of 196m 4. When will it be more than 96m above the ground? Now we need to create an inequalit for S(t)>96m and solve for t: 11t 16t > 96 0 > 16t 11t + 96 move everthing to the same side of the inequalit sign t 7t + 6 < 0 factorise and simplif (t 6)(t 1) < < t < 6 It is above 96 m from 1 to 6 seconds. 5. What is the acceleration of the object? Acceleration is the change in velocit with respect to time. So we need to find the derivative of the velocit: v(t) = 11 3t a(t) = v (t) a(t) = 3 m.s - Activit 7 Activit 1. A large tank is being filled with water through certain pipes. The volume of water in the tank at an given time is given b V(t) = t t, where the volume is measured in cubic metres and the time in minutes. 1.1 Find the rate at which the volume is increasing when t = 1 minute 009 Page 1 Lesson 1 Algebra Page 111

25 1. At what time does the water start running out of the tank?. A cricket ball is thrown verticall up into the air. After seconds, its height is metres where = Determine:.1 the velocit of the ball after 3 seconds. the maimum height reached b the ball..3 the acceleration of the ball..4 the total distance travelled b the ball when it returns to the ground. Activit Activit 8 1. Differentiate the following with respect to : 1.1 ( _ 1 )( + _ 1 ) _ _ 4. Determine: lim ( ) _ 3. If =, determine _ d. Give the answer with positive eponents. d 4. Evaluate: lim _ Determine ƒ () from first principles if ƒ() = 3 6. Find the average gradient of g() between = 1 and = 3 if g() = 7. Determine _ d d if: 7.1 = ( 3 1) _ 7. = _ = ( 1 ) = 7.5 = If ƒ() = _ + _ +, find ƒ () and hence evaluate ƒ (-) Page 11

26 Activit 9 1. The figure shows the curve given b ƒ() = a Point P ( ; 9) lies on the curve of ƒ(). Activit P(; 9) f() 1.1 Determine the value of a. 1. Determine the equation of the tangent at point P. 1.3 Determine the average gradient of the curve between = and = 5. Given: ƒ () = 0 at = 3 and = 1 onl and that ƒ( 5) = 0; ƒ( ) = 0; ƒ(1) = 0; ƒ(0) = 1. Draw a rough sketch of a graph of ƒ() assuming it is a cubic polnomial. 3. ƒ() = b + c + d. A and B are turning points with B(6; 0) 0 B(6; 0) A T f() Find: 3.1 the values of b, c and d. 3. the coordinates of A 3.3 T has an -coordinates of 3, find the -coordinate of T 3.4 Equation of the tangent to the curve at T 4.1 Show that the gradient of the tangent at (; 4) to the curve = _ 4 is double that of the gradient at the point (4; 7) 4. Determine the co-ordinates of the point on the curve at which the gradient is Page 1 Lesson 1 Algebra Page 113

27 Solutions to Activities Activit lim ( + 1) = _ 4 Activit f(6) 8(0) 1.1 gradient = = _ = 6 1. = 3: f (n) = lim f(3 + h) f(3) h 0 h lim (3 + h) + 9 ( 3) + 9) h 0 h lim 9 6h h h 0 h lim h( 6 h) h 0 h 6 0 = 6 (.1 ƒ () = lim + h) + ( + ) h 0 h lim + h + h + h 0 h lim _ h + h h 0 h lim h( + h) h 0 h lim ( + h) = h 0. ƒ () = lim 3( + h) + 3( + h) - (3 + 3) h 0 h 3 lim + 6h + 3h h 3 3 h 0 h lim h(6 + 3h + 3) h 0 h (6 + 3h + 3) = lim h 0 Activt 3 1. ƒ () = _ 1 _ 1 _ 1 _ 3. = 3 ( ) 3. Activit _ d d = = + 1 d _ d = 1 1. f() = + 1 into: (0; 1) f () = gradient of tangent : f (0) = (0) = tangent: = c (sub(0; 1)) = n + 1 Page 114. ƒ() = - + +

28 ƒ () = + 1 at = 1: ƒ (1) = (1) + 1 = 1 \ m of tangent = 1 \ tangent: = + c sub (1 ; ) = 1 + c \ 3 = c \ tangent: = = is a tangent to ƒ() = \ m = 11 and gradient: ƒ () = 4 5 \ 4 5 = 11 4 = 16 \ = 4 to find, sub = -4 into either equation: = 11( 4) + 44 or ƒ( 4) = ( 4) 5( 4) + 1 = 0 = 0 \ point of contact: (-4 ; 0) 4. ƒ() = _ 3 = 3 1 ƒ () = 3 = 3 _ at = -1: f ( 1) = 3 _ ( 1) = 3 _ = 3 1 \ m of tangent = -3 to find the -value: f( 1) = _ 3 1 = 3 \ point of contact: ( 1 ; 3) \ tangent: = 3 + c sub ( 1 ; 3) \ 3 = 3( 1) + c 6 = c \ tangent: = 3 6 Activit 5 1. Finding -intercepts of g(): ( = 0) \ = 0 ( + 1)( 1)( + 1) = 0 \ = 1 or = 1 \ B( 1; 0) and A(1; 0) -intercept of g(): 1 \ C(0; 1) Finding -intercept of f(): ( = 0) 4 + = 0 \ = 1 \ D ( 1 ; 0 ) 009 Page 1 Lesson 1 Algebra Page 115

29 1.1 DA = 1 + _ 1 = 3 OB = 1 CF = 1 + = 3 1. g () = = 0 (3 1)( + 1) = 0 \ = _ 1 3 or = 1 At E, = 1 \ = g 3 ( 1 3) = 3 7 \ E ( 1 3 ; 3 7 ) 1.3 Gradient of f() = 4 (from m = 4) \ g () = = 4 \ = 0 (3 + 5)( 1) = 0 = _ 5 or = 1 (values of for which the gradient of f() and g() are 3 equal.).1 f () = = 0 =,577 or = 1,4 \ =,6 or = 1,4 \ f(,6) = 0,4 f(1,4) = (0,4) (,6; 0,4) (1,4; 0,4). ( 1)( )( 3) = 0 (-intercepts).3 \ = 1 or = or = (1,4; 0,4) ( ; 0) (,6; 0,4) Page = 1; m = = ; m = 1 = 3; m =.5 = _ 1 + c (sub(1; 0)) Note: a normal is a line to the tangent m of tangent = m of normal = _ 1 = _ 1 + _ 1 3. = a ( + 1)( + 1)( 3) sub(0; 9) 9 = a(1)(1)( 3) 9 = 3a a = 3 = 3( + 1)( + 1)( 3) =

30 = No stationar points since derivative = 0 has no solution. However the second derivative equal to zero does have an answer: = 0 so = _ 9 ; = _ Activit 6 1. The numbers are _ 18 5 and 6_ 5 ; the minimum is 14 _ 5. Dimensions are _ 0 cm; 4cm; 10cm 3 _ 3.1 Using Pthag: SR = Note: RC = QC = 30 QR = (30 ) + (30 ) (30 ) _ QR = (30 ) Area PQRS = ( n) ( (30 )) A = (30 ) A = 60 n cm 3. ma Area: A = 60 4 = 0 = 15 ma area A(15) = 60(15) (5) 450 cm Activit Rate of increase: V (t) = 10 t at t = 1: V (1) = 8m 3 /minute 1. ma volume: V (t) = 10 t = 0 t = 5 min.1 velocit = _ d d _ d = d = 3: velocit = = 0 m/s. ma height: _ d = = 0 d = 5 Ma height at = 5: = 50(5) 5(5) = 15m.3 Acceleration = _ d = 10 m/s d.4 Double the maimum = 50 m Activit _ 1 = \ derivative = + _ 3 ( )( 1. 1) = ( ) 1 \ derivative = 1.3 derivative = _ \ derivative = _ 1 = _ _ lim. ( )( + 1) ( ) 009 = lim 1_ ( + 1) = ( + 1) = 6 Page 1 Lesson 1 Algebra Page 117

31 _ 3. = Make the subject of the formula. = _ _ \ = 1 1_ \_ d d = _ _ 1 1_ 4. lim ( )( + + 4) = + () + 4 = 3 ( )( + ) + f( + h) f() = lim 3( + h) ( 3 ) h 0 h h 0 h 3 6h 3h + 3 h 0 h 5. f () = lim = lim = 6 3(0) = 6 g(3) g(1) 6. m = = _ 16 0 = = \_ d d = = + 3 = + 1_ \_ d d = + _ 1 1_ 7.3 = + \_ d d = _ = _ 8 = = _ 1 _ + 1 \_ d d = _ 8 = 1 + \_ d d = _ + _ 3 _ 3 5_ 8. ƒ() = _ \ f () = _ \ ƒ ( ) = _ = _ _ ƒ () = 7 3_ _ Activit sub (; 9) : 9 = a() \ a = 1 1. gradient at = : f () = 3 \ f () = 1 \ tangent: = 1 + c sub (; 9) \ c = 15 and = 1 15 f(5) f() 1.3 m = = _ = 39. Note: f ( 3) = 0 and f ( 1) = 0 \ these are stationar points. f( 5) = 0; f( ) = 0; f(1) = 0 \ these are the -intercepts f(0) = 1 ie. 1 is the -intercept value Page 118

32 3. d = 0 (graph passes through the origin) 3.1 to find b and c ( unknowns) we solve simultaneousl. from B f(6) = 0 \ (6) 3 + 6(6) + c(6) + 0 = 0 6b + c = 36 \ c = 36 6b since B is a stationar point f (6) = 0 : f () = 3 + b + c = 0 f (6) = 3(6) + b(6) + c = 0 (from eqn 1: c = 36 6b) \ b + c = 0 6b = 7 b = 1 \ b b = 0 \ c = eqn: = \_ d d = at A (stationar point) : = = 0 ( 6)( ) = 0 \ A = f() = 3 \ A(; 3) 3.3 f(3) = 7 \ T = m of tangent: f (3) = 3(3) + 4(3) 36 = 9 \ = 9 + c sub T(3; 7) \ C = 54 \ = _ d d = 3 _ 1 gradient at = : 3 _ 1 () = gradient at = 4: 3 _ 1 (4) = 1 \ gradient at (; 4) is double that at point (4; 7) 4.1 (8; 7) 6.1 OP = units; OQ = 1 unit; OR = 6 units; ON = 1 units 6. S(4; 36) 6.3 OT = 30 units; TN = 3 units 6.4 = Z(-1; 14) 7. p = _ (1; s7) 009 Page 1 Lesson 1 Algebra Page 119

Higher. Polynomials and Quadratics 64

Higher. Polynomials and Quadratics 64 hsn.uk.net Higher Mathematics UNIT OUTCOME 1 Polnomials and Quadratics Contents Polnomials and Quadratics 64 1 Quadratics 64 The Discriminant 66 3 Completing the Square 67 4 Sketching Parabolas 70 5 Determining

More information

The Quadratic Function

The Quadratic Function 0 The Quadratic Function TERMINOLOGY Ais of smmetr: A line about which two parts of a graph are smmetrical. One half of the graph is a reflection of the other Coefficient: A constant multiplied b a pronumeral

More information

INVESTIGATIONS AND FUNCTIONS 1.1.1 1.1.4. Example 1

INVESTIGATIONS AND FUNCTIONS 1.1.1 1.1.4. Example 1 Chapter 1 INVESTIGATIONS AND FUNCTIONS 1.1.1 1.1.4 This opening section introduces the students to man of the big ideas of Algebra 2, as well as different was of thinking and various problem solving strategies.

More information

Core Maths C2. Revision Notes

Core Maths C2. Revision Notes Core Maths C Revision Notes November 0 Core Maths C Algebra... Polnomials: +,,,.... Factorising... Long division... Remainder theorem... Factor theorem... 4 Choosing a suitable factor... 5 Cubic equations...

More information

y intercept Gradient Facts Lines that have the same gradient are PARALLEL

y intercept Gradient Facts Lines that have the same gradient are PARALLEL CORE Summar Notes Linear Graphs and Equations = m + c gradient = increase in increase in intercept Gradient Facts Lines that have the same gradient are PARALLEL If lines are PERPENDICULAR then m m = or

More information

If (a)(b) 5 0, then a 5 0 or b 5 0.

If (a)(b) 5 0, then a 5 0 or b 5 0. chapter Algebra Ke words substitution discriminant completing the square real and distinct imaginar rational verte parabola maimum minimum surd irrational rationalising the denominator Section. Quadratic

More information

Solving Quadratic Equations by Graphing. Consider an equation of the form. y ax 2 bx c a 0. In an equation of the form

Solving Quadratic Equations by Graphing. Consider an equation of the form. y ax 2 bx c a 0. In an equation of the form SECTION 11.3 Solving Quadratic Equations b Graphing 11.3 OBJECTIVES 1. Find an ais of smmetr 2. Find a verte 3. Graph a parabola 4. Solve quadratic equations b graphing 5. Solve an application involving

More information

C1: Coordinate geometry of straight lines

C1: Coordinate geometry of straight lines B_Chap0_08-05.qd 5/6/04 0:4 am Page 8 CHAPTER C: Coordinate geometr of straight lines Learning objectives After studing this chapter, ou should be able to: use the language of coordinate geometr find the

More information

SL Calculus Practice Problems

SL Calculus Practice Problems Alei - Desert Academ SL Calculus Practice Problems. The point P (, ) lies on the graph of the curve of = sin ( ). Find the gradient of the tangent to the curve at P. Working:... (Total marks). The diagram

More information

Quadratic Functions. MathsStart. Topic 3

Quadratic Functions. MathsStart. Topic 3 MathsStart (NOTE Feb 2013: This is the old version of MathsStart. New books will be created during 2013 and 2014) Topic 3 Quadratic Functions 8 = 3 2 6 8 ( 2)( 4) ( 3) 2 1 2 4 0 (3, 1) MATHS LEARNING CENTRE

More information

4 Non-Linear relationships

4 Non-Linear relationships NUMBER AND ALGEBRA Non-Linear relationships A Solving quadratic equations B Plotting quadratic relationships C Parabolas and transformations D Sketching parabolas using transformations E Sketching parabolas

More information

Mathematics. Total marks 100. Section I Pages marks Attempt Questions 1 10 Allow about 15 minutes for this section

Mathematics. Total marks 100. Section I Pages marks Attempt Questions 1 10 Allow about 15 minutes for this section 04 HIGHER SCHOOL CERTIFICATE EXAMINATION Mathematics General Instructions Reading time 5 minutes Working time hours Write using black or blue pen Black pen is preferred Board-approved calculators ma be

More information

Section C Non Linear Graphs

Section C Non Linear Graphs 1 of 8 Section C Non Linear Graphs Graphic Calculators will be useful for this topic of 8 Cop into our notes Some words to learn Plot a graph: Draw graph b plotting points Sketch/Draw a graph: Do not plot,

More information

The Not-Formula Book for C1

The Not-Formula Book for C1 Not The Not-Formula Book for C1 Everything you need to know for Core 1 that won t be in the formula book Examination Board: AQA Brief This document is intended as an aid for revision. Although it includes

More information

Coordinate Geometry. Positive gradients: Negative gradients:

Coordinate Geometry. Positive gradients: Negative gradients: 8 Coordinate Geometr Negative gradients: m < 0 Positive gradients: m > 0 Chapter Contents 8:0 The distance between two points 8:0 The midpoint of an interval 8:0 The gradient of a line 8:0 Graphing straight

More information

10.1. Solving Quadratic Equations. Investigation: Rocket Science CONDENSED

10.1. Solving Quadratic Equations. Investigation: Rocket Science CONDENSED CONDENSED L E S S O N 10.1 Solving Quadratic Equations In this lesson you will look at quadratic functions that model projectile motion use tables and graphs to approimate solutions to quadratic equations

More information

Colegio del mundo IB. Programa Diploma REPASO 2. 1. The mass m kg of a radio-active substance at time t hours is given by. m = 4e 0.2t.

Colegio del mundo IB. Programa Diploma REPASO 2. 1. The mass m kg of a radio-active substance at time t hours is given by. m = 4e 0.2t. REPASO. The mass m kg of a radio-active substance at time t hours is given b m = 4e 0.t. Write down the initial mass. The mass is reduced to.5 kg. How long does this take?. The function f is given b f()

More information

To Be or Not To Be a Linear Equation: That Is the Question

To Be or Not To Be a Linear Equation: That Is the Question To Be or Not To Be a Linear Equation: That Is the Question Linear Equation in Two Variables A linear equation in two variables is an equation that can be written in the form A + B C where A and B are not

More information

Functions and Graphs

Functions and Graphs PSf Functions and Graphs Paper 1 Section B 1. The points A and B have coordinates (a, a 2 ) and (2b, 4b 2 ) respectivel. Determine the gradient of AB in its simplest form. 2 2. hsn.uk.net Page 1 Questions

More information

Years t. Definition Anyone who has drawn a circle using a compass will not be surprised by the following definition of the circle: x 2 y 2 r 2 304

Years t. Definition Anyone who has drawn a circle using a compass will not be surprised by the following definition of the circle: x 2 y 2 r 2 304 Section The Circle 65 Dollars Purchase price P Book value = f(t) Salvage value S Useful life L Years t FIGURE 3 Straight-line depreciation. The Circle Definition Anone who has drawn a circle using a compass

More information

HI-RES STILL TO BE SUPPLIED

HI-RES STILL TO BE SUPPLIED 1 MRE GRAPHS AND EQUATINS HI-RES STILL T BE SUPPLIED Different-shaped curves are seen in man areas of mathematics, science, engineering and the social sciences. For eample, Galileo showed that if an object

More information

Alex and Morgan were asked to graph the equation y = 2x + 1

Alex and Morgan were asked to graph the equation y = 2x + 1 Which is better? Ale and Morgan were asked to graph the equation = 2 + 1 Ale s make a table of values wa Morgan s use the slope and -intercept wa First, I made a table. I chose some -values, then plugged

More information

8.7 Systems of Non-Linear Equations and Inequalities

8.7 Systems of Non-Linear Equations and Inequalities 8.7 Sstems of Non-Linear Equations and Inequalities 67 8.7 Sstems of Non-Linear Equations and Inequalities In this section, we stud sstems of non-linear equations and inequalities. Unlike the sstems of

More information

SAMPLE. Polynomial functions

SAMPLE. Polynomial functions Objectives C H A P T E R 4 Polnomial functions To be able to use the technique of equating coefficients. To introduce the functions of the form f () = a( + h) n + k and to sketch graphs of this form through

More information

Chapter 6 Quadratic Functions

Chapter 6 Quadratic Functions Chapter 6 Quadratic Functions Determine the characteristics of quadratic functions Sketch Quadratics Solve problems modelled b Quadratics 6.1Quadratic Functions A quadratic function is of the form where

More information

Zeros of Polynomial Functions. The Fundamental Theorem of Algebra. The Fundamental Theorem of Algebra. zero in the complex number system.

Zeros of Polynomial Functions. The Fundamental Theorem of Algebra. The Fundamental Theorem of Algebra. zero in the complex number system. _.qd /7/ 9:6 AM Page 69 Section. Zeros of Polnomial Functions 69. Zeros of Polnomial Functions What ou should learn Use the Fundamental Theorem of Algebra to determine the number of zeros of polnomial

More information

x 2. 4x x 4x 1 x² = 0 x = 0 There is only one x-intercept. (There can never be more than one y-intercept; do you know why?)

x 2. 4x x 4x 1 x² = 0 x = 0 There is only one x-intercept. (There can never be more than one y-intercept; do you know why?) Math Learning Centre Curve Sketching A good graphing calculator can show ou the shape of a graph, but it doesn t alwas give ou all the useful information about a function, such as its critical points and

More information

t hours This is the distance in miles travelled in 2 hours when the speed is 70mph. = 22 yards per second. = 110 yards.

t hours This is the distance in miles travelled in 2 hours when the speed is 70mph. = 22 yards per second. = 110 yards. The area under a graph often gives useful information. Velocit-time graphs Constant velocit The sketch shows the velocit-time graph for a car that is travelling along a motorwa at a stead 7 mph. 7 The

More information

Course 2 Answer Key. 1.1 Rational & Irrational Numbers. Defining Real Numbers Student Logbook. The Square Root Function Student Logbook

Course 2 Answer Key. 1.1 Rational & Irrational Numbers. Defining Real Numbers Student Logbook. The Square Root Function Student Logbook Course Answer Ke. Rational & Irrational Numbers Defining Real Numbers. integers; 0. terminates; repeats 3. two; number 4. ratio; integers 5. terminating; repeating 6. rational; irrational 7. real 8. root

More information

13 Graphs, Equations and Inequalities

13 Graphs, Equations and Inequalities 13 Graphs, Equations and Inequalities 13.1 Linear Inequalities In this section we look at how to solve linear inequalities and illustrate their solutions using a number line. When using a number line,

More information

2.4 Inequalities with Absolute Value and Quadratic Functions

2.4 Inequalities with Absolute Value and Quadratic Functions 08 Linear and Quadratic Functions. Inequalities with Absolute Value and Quadratic Functions In this section, not onl do we develop techniques for solving various classes of inequalities analticall, we

More information

6. The given function is only drawn for x > 0. Complete the function for x < 0 with the following conditions:

6. The given function is only drawn for x > 0. Complete the function for x < 0 with the following conditions: Precalculus Worksheet 1. Da 1 1. The relation described b the set of points {(-, 5 ),( 0, 5 ),(,8 ),(, 9) } is NOT a function. Eplain wh. For questions - 4, use the graph at the right.. Eplain wh the graph

More information

Algebra 2 Honors: Quadratic Functions. Student Focus

Algebra 2 Honors: Quadratic Functions. Student Focus Resources: SpringBoard- Algebra Online Resources: Algebra Springboard Tet Algebra Honors: Quadratic Functions Semester 1, Unit : Activit 10 Unit Overview In this unit, students write the equations of quadratic

More information

Quadratic Functions and Models. The Graph of a Quadratic Function. These functions are examples of polynomial functions. Why you should learn it

Quadratic Functions and Models. The Graph of a Quadratic Function. These functions are examples of polynomial functions. Why you should learn it 0_00.qd 8 /7/05 Chapter. 9:0 AM Page 8 Polnomial and Rational Functions Quadratic Functions and Models What ou should learn Analze graphs of quadratic functions. Write quadratic functions in standard form

More information

Quadratic Functions. Academic Vocabulary justify derive verify

Quadratic Functions. Academic Vocabulary justify derive verify Quadratic Functions 015 College Board. All rights reserved. Unit Overview This unit focuses on quadratic functions and equations. You will write the equations of quadratic functions to model situations.

More information

Exponential and Logarithmic Functions

Exponential and Logarithmic Functions Chapter 6 Eponential and Logarithmic Functions Section summaries Section 6.1 Composite Functions Some functions are constructed in several steps, where each of the individual steps is a function. For eample,

More information

MATHEMATICS. Prelim Revision. (with answers)

MATHEMATICS. Prelim Revision. (with answers) MATHEMATICS Prelim Revision (with answers) FRMULAE LIST The roots of b b 4ac a b c 0 are a Sine Rule: a sina b sin c sinc Cosine Rule: a b c b c a bccos A or cos A bc Area of a triangle: 1 A absinc Volume

More information

2 Analysis of Graphs of

2 Analysis of Graphs of ch.pgs1-16 1/3/1 1:4 AM Page 1 Analsis of Graphs of Functions A FIGURE HAS rotational smmetr around an ais I if it coincides with itself b all rotations about I. Because of their complete rotational smmetr,

More information

Review of Essential Skills and Knowledge

Review of Essential Skills and Knowledge Review of Essential Skills and Knowledge R Eponent Laws...50 R Epanding and Simplifing Polnomial Epressions...5 R 3 Factoring Polnomial Epressions...5 R Working with Rational Epressions...55 R 5 Slope

More information

A Resource for Free-standing Mathematics Qualifications

A Resource for Free-standing Mathematics Qualifications To find a maximum or minimum: Find an expression for the quantity you are trying to maximise/minimise (y say) in terms of one other variable (x). dy Find an expression for and put it equal to 0. Solve

More information

STRAND F: ALGEBRA. UNIT F4 Solving Quadratic Equations: Text * * * Contents. Section. F4.1 Factorisation. F4.2 Using the Formula

STRAND F: ALGEBRA. UNIT F4 Solving Quadratic Equations: Text * * * Contents. Section. F4.1 Factorisation. F4.2 Using the Formula UNIT F4 Solving Quadratic Equations: Tet STRAND F: ALGEBRA Unit F4 Solving Quadratic Equations Tet Contents * * * Section F4. Factorisation F4. Using the Formula F4. Completing the Square UNIT F4 Solving

More information

Graphing and transforming functions

Graphing and transforming functions Chapter 5 Graphing and transforming functions Contents: A B C D Families of functions Transformations of graphs Simple rational functions Further graphical transformations Review set 5A Review set 5B 6

More information

Graphing Quadratic Equations

Graphing Quadratic Equations .4 Graphing Quadratic Equations.4 OBJECTIVE. Graph a quadratic equation b plotting points In Section 6.3 ou learned to graph first-degree equations. Similar methods will allow ou to graph quadratic equations

More information

The only idea of real

The only idea of real 80 Chapter Polnomial and Rational Functions. RATIONAL FUNCTIONS What is mathematics about? I think it s reall summed up in what I frequentl tell m classes. That is that proofs reall aren t there to convince

More information

Core Maths C3. Revision Notes

Core Maths C3. Revision Notes Core Maths C Revision Notes October 0 Core Maths C Algebraic fractions... Cancelling common factors... Multipling and dividing fractions... Adding and subtracting fractions... Equations... 4 Functions...

More information

Are You Ready? Circumference and Area of Circles

Are You Ready? Circumference and Area of Circles SKILL 39 Are You Read? Circumference and Area of Circles Teaching Skill 39 Objective Find the circumference and area of circles. Remind students that perimeter is the distance around a figure and that

More information

STRAND: ALGEBRA Unit 3 Solving Equations

STRAND: ALGEBRA Unit 3 Solving Equations CMM Subject Support Strand: ALGEBRA Unit Solving Equations: Tet STRAND: ALGEBRA Unit Solving Equations TEXT Contents Section. Algebraic Fractions. Algebraic Fractions and Quadratic Equations. Algebraic

More information

Math 1400/1650 (Cherry): Quadratic Functions

Math 1400/1650 (Cherry): Quadratic Functions Math 100/1650 (Cherr): Quadratic Functions A quadratic function is a function Q() of the form Q() = a + b + c with a 0. For eample, Q() = 3 7 + 5 is a quadratic function, and here a = 3, b = 7 and c =

More information

Implicit Differentiation

Implicit Differentiation Revision Notes 2 Calculus 1270 Fall 2007 INSTRUCTOR: Peter Roper OFFICE: LCB 313 [EMAIL: roper@math.utah.edu] Standard Disclaimer These notes are not a complete review of the course thus far, and some

More information

Mathematics Paper 1 (Non-Calculator)

Mathematics Paper 1 (Non-Calculator) H National Qualifications CFE Higher Mathematics - Specimen Paper A Duration hour and 0 minutes Mathematics Paper (Non-Calculator) Total marks 60 Attempt ALL questions. You ma NOT use a calculator. Full

More information

17.1 Connecting Intercepts and Zeros

17.1 Connecting Intercepts and Zeros Locker LESSON 7. Connecting Intercepts and Zeros Teas Math Standards The student is epected to: A.7.A Graph quadratic functions on the coordinate plane and use the graph to identif ke attributes, if possible,

More information

Practice for Final Disclaimer: The actual exam is not a mirror of this. These questions are merely an aid to help you practice 1 2)

Practice for Final Disclaimer: The actual exam is not a mirror of this. These questions are merely an aid to help you practice 1 2) Practice for Final Disclaimer: The actual eam is not a mirror of this. These questions are merel an aid to help ou practice Solve the problem. 1) If m varies directl as p, and m = 7 when p = 9, find m

More information

POLYNOMIAL FUNCTIONS

POLYNOMIAL FUNCTIONS POLYNOMIAL FUNCTIONS Polynomial Division.. 314 The Rational Zero Test.....317 Descarte s Rule of Signs... 319 The Remainder Theorem.....31 Finding all Zeros of a Polynomial Function.......33 Writing a

More information

When I was 3.1 POLYNOMIAL FUNCTIONS

When I was 3.1 POLYNOMIAL FUNCTIONS 146 Chapter 3 Polnomial and Rational Functions Section 3.1 begins with basic definitions and graphical concepts and gives an overview of ke properties of polnomial functions. In Sections 3.2 and 3.3 we

More information

Applications of Quadratic Functions Word Problems ISU. Part A: Revenue and Numeric Problems

Applications of Quadratic Functions Word Problems ISU. Part A: Revenue and Numeric Problems Mr. Gardner s MPMD Applications of Quadratic Functions Word Problems ISU Part A: Revenue and Numeric Problems When ou solve problems using equations, our solution must have four components: 1. A let statement,

More information

Contents. How You May Use This Resource Guide

Contents. How You May Use This Resource Guide Contents How You Ma Use This Resource Guide ii 9 Fractional and Quadratic Equations 1 Worksheet 9.1: Similar Figures.......................... 5 Worksheet 9.: Stretch of a Spring........................

More information

1.6 Graphs of Functions

1.6 Graphs of Functions .6 Graphs of Functions 9.6 Graphs of Functions In Section. we defined a function as a special tpe of relation; one in which each -coordinate was matched with onl one -coordinate. We spent most of our time

More information

Polynomial and Rational Functions

Polynomial and Rational Functions Chapter 5 Polnomial and Rational Functions Section 5.1 Polnomial Functions Section summaries The general form of a polnomial function is f() = a n n + a n 1 n 1 + +a 1 + a 0. The degree of f() is the largest

More information

8. Bilateral symmetry

8. Bilateral symmetry . Bilateral smmetr Our purpose here is to investigate the notion of bilateral smmetr both geometricall and algebraicall. Actuall there's another absolutel huge idea that makes an appearance here and that's

More information

Graphing Quadratic Functions

Graphing Quadratic Functions A. THE STANDARD PARABOLA Graphing Quadratic Functions The graph of a quadratic function is called a parabola. The most basic graph is of the function =, as shown in Figure, and it is to this graph which

More information

Polynomials Past Papers Unit 2 Outcome 1

Polynomials Past Papers Unit 2 Outcome 1 PSf Polnomials Past Papers Unit 2 utcome 1 Multiple Choice Questions Each correct answer in this section is worth two marks. 1. Given p() = 2 + 6, which of the following are true? I. ( + 3) is a factor

More information

DISTANCE, CIRCLES, AND QUADRATIC EQUATIONS

DISTANCE, CIRCLES, AND QUADRATIC EQUATIONS a p p e n d i g DISTANCE, CIRCLES, AND QUADRATIC EQUATIONS DISTANCE BETWEEN TWO POINTS IN THE PLANE Suppose that we are interested in finding the distance d between two points P (, ) and P (, ) in the

More information

Polynomial Degree and Finite Differences

Polynomial Degree and Finite Differences CONDENSED LESSON 7.1 Polynomial Degree and Finite Differences In this lesson you will learn the terminology associated with polynomials use the finite differences method to determine the degree of a polynomial

More information

LESSON EIII.E EXPONENTS AND LOGARITHMS

LESSON EIII.E EXPONENTS AND LOGARITHMS LESSON EIII.E EXPONENTS AND LOGARITHMS LESSON EIII.E EXPONENTS AND LOGARITHMS OVERVIEW Here s what ou ll learn in this lesson: Eponential Functions a. Graphing eponential functions b. Applications of eponential

More information

Higher. Functions and Graphs. Functions and Graphs 18

Higher. Functions and Graphs. Functions and Graphs 18 hsn.uk.net Higher Mathematics UNIT UTCME Functions and Graphs Contents Functions and Graphs 8 Sets 8 Functions 9 Composite Functions 4 Inverse Functions 5 Eponential Functions 4 6 Introduction to Logarithms

More information

Conics and Polar Coordinates

Conics and Polar Coordinates Contents 17 Conics and Polar Coordinates 17.1 Conic Sections 2 17.2 Polar Coordinates 23 17.3 Parametric Curves 33 Learning outcomes In this Workbook ou will learn about some of the most important curves

More information

Chapter 3. Curve Sketching. By the end of this chapter, you will

Chapter 3. Curve Sketching. By the end of this chapter, you will Chapter 3 Curve Sketching How much metal would be required to make a -ml soup can? What is the least amount of cardboard needed to build a bo that holds 3 cm 3 of cereal? The answers to questions like

More information

The first derivative and stationary points

The first derivative and stationary points Mathematics Learning Centre The first derivative and stationar points Jackie Nicholas c 24 Universit of Sdne Mathematics Learning Centre, Universit of Sdne The first derivative and stationar points The

More information

Algebra II Quadratic Functions and Equations- Transformations Unit 05a

Algebra II Quadratic Functions and Equations- Transformations Unit 05a Previous Knowledge: (What skills do the need to have to succeed?) Squares and Square Roots Simplif Radical Expressions Multipl Binomials Solve Multi-Step Equations Identif and graph linear functions Transform

More information

Functions. Chapter A B C D E F G H I J. The reciprocal function x

Functions. Chapter A B C D E F G H I J. The reciprocal function x Chapter 1 Functions Contents: A B C D E F G H I J Relations and functions Function notation, domain and range Composite functions, f± g Sign diagrams Inequalities (inequations) The modulus function 1 The

More information

Section P.9 Notes Page 1 P.9 Linear Inequalities and Absolute Value Inequalities

Section P.9 Notes Page 1 P.9 Linear Inequalities and Absolute Value Inequalities Section P.9 Notes Page P.9 Linear Inequalities and Absolute Value Inequalities Sometimes the answer to certain math problems is not just a single answer. Sometimes a range of answers might be the answer.

More information

Section 7.2 Linear Programming: The Graphical Method

Section 7.2 Linear Programming: The Graphical Method Section 7.2 Linear Programming: The Graphical Method Man problems in business, science, and economics involve finding the optimal value of a function (for instance, the maimum value of the profit function

More information

Introduction. Introduction

Introduction. Introduction Introduction Solving Sstems of Equations Let s start with an eample. Recall the application of sales forecasting from the Working with Linear Equations module. We used historical data to derive the equation

More information

Core Maths C1. Revision Notes

Core Maths C1. Revision Notes Core Maths C Revision Notes November 0 Core Maths C Algebra... Indices... Rules of indices... Surds... 4 Simplifying surds... 4 Rationalising the denominator... 4 Quadratic functions... 4 Completing the

More information

C3: Functions. Learning objectives

C3: Functions. Learning objectives CHAPTER C3: Functions Learning objectives After studing this chapter ou should: be familiar with the terms one-one and man-one mappings understand the terms domain and range for a mapping understand the

More information

3.1 Quadratic Functions

3.1 Quadratic Functions 33337_030.qp 252 2/27/06 Chapter 3 :20 PM Page 252 Polnomial and Rational Functions 3. Quadratic Functions The Graph of a Quadratic Function In this and the net section, ou will stud the graphs of polnomial

More information

Polynomial and Rational Functions

Polynomial and Rational Functions Chapter Section.1 Quadratic Functions Polnomial and Rational Functions Objective: In this lesson ou learned how to sketch and analze graphs of quadratic functions. Course Number Instructor Date Important

More information

SECTION 2-5 Combining Functions

SECTION 2-5 Combining Functions 2- Combining Functions 16 91. Phsics. A stunt driver is planning to jump a motorccle from one ramp to another as illustrated in the figure. The ramps are 10 feet high, and the distance between the ramps

More information

Quadratic Functions. Unit

Quadratic Functions. Unit Quadratic Functions Unit 5 Unit Overview In this unit ou will stud a variet of was to solve quadratic functions and appl our learning to analzing real world problems. Academic Vocabular Add these words

More information

Quadratic Equations and Functions

Quadratic Equations and Functions Quadratic Equations and Functions. Square Root Propert and Completing the Square. Quadratic Formula. Equations in Quadratic Form. Graphs of Quadratic Functions. Verte of a Parabola and Applications In

More information

Click here for answers.

Click here for answers. CHALLENGE PROBLEMS: CHALLENGE PROBLEMS 1 CHAPTER A Click here for answers S Click here for solutions A 1 Find points P and Q on the parabola 1 so that the triangle ABC formed b the -ais and the tangent

More information

Investigation. So far you have worked with quadratic equations in vertex form and general. Getting to the Root of the Matter. LESSON 9.

Investigation. So far you have worked with quadratic equations in vertex form and general. Getting to the Root of the Matter. LESSON 9. DA2SE_73_09.qd 0/8/0 :3 Page Factored Form LESSON 9.4 So far you have worked with quadratic equations in verte form and general form. This lesson will introduce you to another form of quadratic equation,

More information

Practice Unit tests Use this booklet to help you prepare for all unit tests in Higher Maths.

Practice Unit tests Use this booklet to help you prepare for all unit tests in Higher Maths. Practice Unit tests Use this booklet to help you prepare for all unit tests in Higher Maths. Your formal test will be of a similar standard. Read the description of each assessment standard carefully to

More information

College Algebra - MAT 161 Page: 1 Copyright 2009 Killoran

College Algebra - MAT 161 Page: 1 Copyright 2009 Killoran College Algera - MAT 161 Page: 1 Copright 009 Killoran Quadratic Functions The graph of f./ D a C C c (where a,,c are real and a 6D 0) is called a paraola. Paraola s are Smmetric over the line that passes

More information

Rational Functions. 7.1 A Rational Existence. 7.2 A Rational Shift in Behavior. 7.3 A Rational Approach. 7.4 There s a Hole In My Function, Dear Liza

Rational Functions. 7.1 A Rational Existence. 7.2 A Rational Shift in Behavior. 7.3 A Rational Approach. 7.4 There s a Hole In My Function, Dear Liza Rational Functions 7 The ozone laer protects Earth from harmful ultraviolet radiation. Each ear, this laer thins dramaticall over the poles, creating ozone holes which have stretched as far as Australia

More information

2.2 Absolute Value Functions

2.2 Absolute Value Functions . Absolute Value Functions 7. Absolute Value Functions There are a few was to describe what is meant b the absolute value of a real number. You ma have been taught that is the distance from the real number

More information

4-1. Quadratic Functions and Transformations. Vocabulary. Review. Vocabulary Builder. Use Your Vocabulary

4-1. Quadratic Functions and Transformations. Vocabulary. Review. Vocabulary Builder. Use Your Vocabulary 4-1 Quadratic Functions and Transformations Vocabular Review 1. Circle the verte of each absolute value graph. Vocabular Builder parabola (noun) puh RAB uh luh Related Words: verte, ais of smmetr, quadratic

More information

Polynomials. Jackie Nicholas Jacquie Hargreaves Janet Hunter

Polynomials. Jackie Nicholas Jacquie Hargreaves Janet Hunter Mathematics Learning Centre Polnomials Jackie Nicholas Jacquie Hargreaves Janet Hunter c 26 Universit of Sdne Mathematics Learning Centre, Universit of Sdne 1 1 Polnomials Man of the functions we will

More information

Quadratic Equations in One Unknown

Quadratic Equations in One Unknown 1 Quadratic Equations in One Unknown 1A 1. Solving Quadratic Equations Using the Factor Method Name : Date : Mark : Ke Concepts and Formulae 1. An equation in the form a + b + c, where a, b and c are real

More information

Families of Quadratics

Families of Quadratics Families of Quadratics Objectives To understand the effects of a, b, and c on the graphs of parabolas of the form a 2 b c To use quadratic equations and graphs to analze the motion of projectiles To distinguish

More information

THE PARABOLA section. Developing the Equation

THE PARABOLA section. Developing the Equation 80 (-0) Chapter Nonlinear Sstems and the Conic Sections. THE PARABOLA In this section Developing the Equation Identifing the Verte from Standard Form Smmetr and Intercepts Graphing a Parabola Maimum or

More information

PROPERTIES OF ELLIPTIC CURVES AND THEIR USE IN FACTORING LARGE NUMBERS

PROPERTIES OF ELLIPTIC CURVES AND THEIR USE IN FACTORING LARGE NUMBERS PROPERTIES OF ELLIPTIC CURVES AND THEIR USE IN FACTORING LARGE NUMBERS A ver important set of curves which has received considerabl attention in recent ears in connection with the factoring of large numbers

More information

1. (a) Find an equation of the line joining A (7, 4) and B (2, 0), giving your answer in the form ax + by + c = 0, where a, b and c are integers.

1. (a) Find an equation of the line joining A (7, 4) and B (2, 0), giving your answer in the form ax + by + c = 0, where a, b and c are integers. 1. (a) Find an equation of the line joining A (7, 4) and B (2, 0), giving your answer in the form a + by + c = 0, where a, b and c are integers. (b) Find the length of AB, leaving your answer in surd form.

More information

Equation of a Line. Chapter H2. The Gradient of a Line. m AB = Exercise H2 1

Equation of a Line. Chapter H2. The Gradient of a Line. m AB = Exercise H2 1 Chapter H2 Equation of a Line The Gradient of a Line The gradient of a line is simpl a measure of how steep the line is. It is defined as follows :- gradient = vertical horizontal horizontal A B vertical

More information

AQA Level 2 Certificate FURTHER MATHEMATICS

AQA Level 2 Certificate FURTHER MATHEMATICS AQA Qualifications AQA Level 2 Certificate FURTHER MATHEMATICS Level 2 (8360) Our specification is published on our website (www.aqa.org.uk). We will let centres know in writing about any changes to the

More information

3 Functions and Graphs

3 Functions and Graphs 54617_CH03_155-224.QXP 9/14/10 1:04 PM Page 155 3 Functions and Graphs In This Chapter 3.1 Functions and Graphs 3.2 Smmetr and Transformations 3.3 Linear and Quadratic Functions 3.4 Piecewise-Defined Functions

More information

Chapter 7: Eigenvalues and Eigenvectors 16

Chapter 7: Eigenvalues and Eigenvectors 16 Chapter 7: Eigenvalues and Eigenvectors 6 SECION G Sketching Conics B the end of this section ou will be able to recognise equations of different tpes of conics complete the square and use this to sketch

More information

g = l 2π g = bd b c d = ac bd. Therefore to write x x and as a single fraction we do the following

g = l 2π g = bd b c d = ac bd. Therefore to write x x and as a single fraction we do the following OCR Core 1 Module Revision Sheet The C1 exam is 1 hour 30 minutes long. You are not allowed any calculator 1. Before you go into the exam make sureyou are fully aware of the contents of theformula booklet

More information

To find a maximum or minimum: Find an expression for the quantity you are trying to maximise/minimise (y, say) in terms of one other variable (x).

To find a maximum or minimum: Find an expression for the quantity you are trying to maximise/minimise (y, say) in terms of one other variable (x). Maxima and minima In this activity you will learn how to use differentiation to find maximum and minimum values of functions. You will then put this into practice on functions that model practical contexts.

More information

Lesson 6: Linear Functions and their Slope

Lesson 6: Linear Functions and their Slope Lesson 6: Linear Functions and their Slope A linear function is represented b a line when graph, and represented in an where the variables have no whole number eponent higher than. Forms of a Linear Equation

More information