Key Terms: Quadratic function. Parabola. Vertex (of a parabola) Minimum value. Maximum value. Axis of symmetry. Vertex form (of a quadratic function)

Transcription

1 Outcome R3 Quadratic Functions McGraw-Hill 3.1, 3.2 Key Terms: Quadratic function Parabola Vertex (of a parabola) Minimum value Maximum value Axis of symmetry Vertex form (of a quadratic function) Standard form (of a quadratic function) p. 1 of 16

2 Lesson 3.1 a: Investigating Quadratic Functions THINK: You are given the following set of squares with various side lengths Create a table of values and graph the perimeter of each square: Side length (x) Perimeter (y) 1 unit 4 units 7 units Create a table of values and graph the area of each square: Side length (x) Area (y) 1 unit 1 square unit 7 units How are these two graphs the same? How are they different? p. 2 of 16

3 The above two graphs are the same in that they both deal with squares of different lengths. One graph shows the perimeter of a square as a straight line. The other graph shows the area of a square which we see as increasing at a faster rate. How do we compare these two graphs mathematically? - The perimeter graph shows a function. This means that the biggest exponent on the is 1. We don t see this 1, it just looks like an. Ex. = The area graph shows a function. This means that there is an of in the equation. Ex. =, = 3 4 General Form of a Quadratic Equation: 2 y = ax + bx + c (2 nd degree equation) - where, and are real numbers and 0. - This is called the general form of the equation of a quadratic function Examples: Graphing: Graph the following equations by: (i) creating a table of values, (ii) plotting the coordinates, and (iii) joining the points with a smooth curve. Ex.1 2 y = x x y The shape of the graph of a quadratic function is called a p. 3 of 16

4 Ex. 2 y = ( x 3) 2 7 x y The characteristics of a parabola are as follows: i. The of a parabola is its highest or lowest point. The vertex may be a or a ii. iii. The bisects the parabola at the vertex. The parabola is reflected about this line. When the coefficient of is, the parabola and its vertex is a minimum point. When the coefficient of is, the parabola and its vertex is a maximum point. Positive: Negative: iv. The of a quadratic function is all possible x-values. The domain of all parabolas is R or v. The of a quadratic function is all possible y-values. It depends on whether the parabola up (goes to positive ) or down (goes to negative ). Also, it depends on the position of the -value of the vertex. p. 4 of 16

5 Example: For the following graphs of quadratic functions, state the following characteristics: Vertex: Graph #1 = Axis of Symmetry: Max or Min? at: Domain: Range: x-intercept(s): y-intercept: Solution: i) To find the points to plot, substitute each value of in = , then determine the corresponding value of. x y ii) From the graph, the -intercepts are -1 and 3. The -intercept is 6. - From the graph, the coordinates of the vertex are the coordinates of the maximum point. They appear to be: - Draw a vertical dotted line through the vertex. The line passes through 1 on the -axis. So, the equation of the axes of symmetry is : - The domain is all possible -values. The domain is - The range is all possible -values. The greatest -value is the -coordinate of the vertex. So, the range is: p. 5 of 16

6 If the -intercepts cannot be identified from the table or graph, they can be determined by solving the related quadratic equation; o that is, the -intercepts are the values of when =0. For Graph #1, solve 0= Divide each term by the common factor 2 Solve by factoring The x-intercepts of the graph of a quadratic function = + + is where the curve crosses the x-axis, this is when = 0. The solution to the equation 0 = + + are the of the quadratic. The zeroes of the function are the for which ()=0. This is the solution of the related quadratic equation + + = 0. Graph #2 Try it on your own: = Vertex: Axis of Symmetry: Max or Min? at: Domain: Range: x-intercept(s): y-intercept: p. 6 of 16

7 Lesson 3.1 b: Quadratic Functions in Vertex form Ok, so now that we have seen the basic make up of a quadratic function, let s look at this specific form of a quadratic from which we can see the value of the vertex. = ( h) + The number in this position determines direction of opening and width of the parabola. The number in this position is the x value of the vertex The number in this position is the y value of the vertex. 1) The effect of changing the a value in = : The parameter determines the direction of opening and the width of the graph. This is known as a vertical stretch. If >, the graph opens. If < the graph opens. This can also be thought of as a reflection about the. If the value of between -1 and 1 or -1<a<1, then the parabola appears wider. If the value of >" or <", the parabola appears to be. p. 7 of 16

8 Sketch and compare the graphs below on the same coordinate grid. = = = " In words, describe the effect of the parameter on the graphs. Now compare to the sketches of = = = " p. 8 of 16

9 2) Changing the value of k in = +# The graph of = + is the image of the graph of = after a translation of units. When k is the graph shifts. When k is the graph shifts. Example 1: Sketch i) = the parent function. ii) = +2 the parent function shifted 2 units up. iii) = 3 the parent function shifted 3 units down. p. 9 of 16

10 3) Changing the value of h in = ( h) The graph of = ( h) is the graph of = () after a of h units. **Note: The value of h is to the connecting sign. Example: For = ( 3), the value of h= 3. Why? Because ( (+3)) For = (+2), the value of h= 2. Why? Because ( ( 2)) Sketch the following using the horizontal shifts. = () the parent function. = ( 3) the parent function shifted 3 units to the right. = (+4) the parent function shifted 4 units to the left. p. 10 of 16

11 Example 5: Given the following equations, determine the following and sketch the graph. 2 a) y = 3( x + 1) + 12 Vertex Axis of symmetry Direction of opening y-intercept Domain Range Max/min value Vertical stretch factor Zero(s) x-intercept b) 2 ( 2) 2 x 1 y = + 3 Vertex Axis of symmetry Direction of opening y-intercept Domain Range Max/min value Vertical stretch factor Zero(s) x-intercept p. 11 of 16

12 2 c) y = 2( x 3) 5 Vertex Axis of symmetry Direction of opening y-intercept Domain Range Max/min value Vertical stretch factor Zero(s) x-intercept Note: Always 1. stretch and reflect the graph first, then 2. translate vertically or horizontally. Example 6: Find the equation of the quadratic function whose vertex is (4, -7) and that passes through the point (2, 9). Homework: McGraw-Hill Read page 156 Key ideas P. 157 # 1-4, 6 (Basic) # 7-9 (Intermediate) # (Advanced) p. 12 of 16

13 Lesson 3.2 : Quadratic Functions in Standard form = ++ a determines the direction of opening and the shape of the graph, a 0 (just as in the vertex form). b influences the position of the graph. c determines the y-intercept of the graph. This is because x=0 at that point. The x value of the vertex is = $ % Let s expand this vertex form &()= " ' ( ) ( to get the standard form of the quadratic. What are the advantages of each form? Check out the factored form of the following quadratics, can you determine why the x- value of the vertex is = $ %? How would you find the y-value of the vertex? Example 1: y-int = +4 5 y-int = 5 6 x-int x-int AoS AoS Vertex Vertex p. 13 of 16

14 Determining the Equation Given Characteristics of the Graph Factored form: =( " )( ) In factored form * and are the +,-./.0-1. Example 2: The graph of a quadratic function passes through B(3,16), and the zeros of the function are 1 and 7. Write the equation of the graph in Standard form. Solution: The zeros of the function are the x-intercepts of the graph, so use the factored form of the equation: =( " )( ) Substitute x 1 = and x 2 = The equation becomes = The coordinates of B(3,16) satisfy the equation, so substitute these into the equation to determine the value of a. Use x=3 and y=16. =( ")( 2) In factored form, the equation is: = In Standard Form: p. 14 of 16

15 Example 3: A bridge spans a horizontal distance of 40m and has a parabolic arch. One metre from the end of the bridge, the arch is 1.95m high. a) Determine an equation that represents this parabolic arch. b) How high is the arch at the centre of the bridge? Solution: a) Sketch the parabola with one end of the arch at the origin(, ). Since the arch spans, the other end of the arch has coordinates (,0). (1, 1.95) (0,0) (40,0) The x-intercepts are and. Factored form of the equation: =( )( ) The coordinates of the given point satisfy the equation substitute and solve for a. =( )( ' ) The equation is: = p. 15 of 16

16 Solution: b) The centre of the arch is at the. The height of the arch is the y-coordinate of the vertex. The x-coordinate of the vertex is halfway between and ; which is. Substitute in to the equation: = The arch is metres high. This next example is not on the video Try it! Example 4: You would like to build a rectangular fence to maximize the area inside. You have enough money for 160m of fencing. What are the dimensions of the fence that you build? a) Write a quadratic function in standard form to represent the area inside your fence. Perimeter of fence = b) Determine the vertex. What does this represent? Homework: McGraw-Hill P. 174 #1-4, 6-9 (Basic) #10-14 (Intermediate) #19, (Advanced) p. 16 of 16