AP Calculus I Summer Packet

Transcription

1 AP Calculus I Summer Packet You will need to use your own paper to work the problems in this packet. You will turn in ALL of your work and the attached completed answer sheet. I. Intercepts The -intercept is where the graph crosses the -ais. You can find the -intercept by setting y=0. The y-intercept is where the graph crosses the y-ais. You can find the y-intercept by setting =0. Eample: Find the intercepts for y ( ) 4 -intercept 0 ( ) 4 4 set y=0 ( ) add 4 to both sides ( ) take square root of both sides ( ) or ( ) Write as equations or Subtract from both sides y-intercept y (0 ) 4 set =0 y 4 Add 0+ y 9 4 Square y subtract Problem Set I Find the intercepts for each of the following.. y 6 4. y 4. 0 y 4. y 6. y 6

2 II. Lines The slope intercept form of a line is y=m+b where m is the slope and b is the y-intercept. The point slope form of a line is y y m( ) where m is the slope and ) is a point on the line. If two lines are parallel then they have the same slope. If two ( y, lines are perpendicular then they have negative reciprocal slopes. Eample: Find the slope of the lines parallel and perpendicular to y The slope of this line is m The parallel line has slope m and the perpendicular line has slope m. Eample: Find the equations of (a) line parallel and (b) perpendicular to y that contains the point (-,) Part a (using slope from eample above) y = ( + ) Using the point-slope form with m and point (-,) Part b (using slope from eample above) y = ( + ) Using the point-slope form with m and point (-,) Eample: Find the slope and y-intercept of 6 y First you must get the line in slope-intercept form. y 6 Subtract 6 form both sides 6 Divide by - y 6 Simplify y The slope is m= 6 and the y-intercept is - Eample: Find the equation of the line that passes through (-,) and (4,). You will need to find slope using m y y m choose one point to substitute back into either the point slope or slope-intercept form of a line. 4 y = ( 4) Using the point-slope form with m and point (4,) Problem Set II Find the equation of a line:. contains (,-4) and (,). contains,, 4 6. contains (-,4) and m is undefined 4. contains (-,-) and m=. contains (-,4) and m 6. contains (0,0) and (,6) 7. -intercept (,0) and y-intercept (0,) Find the slope and y-intercept of the line: 8. +y = 0 Sketch a graph of the equation: 9. y=- 0. =4. y-=+ Write an equation of a line through the point (a). parallel to the given line and (b) perpendicular to the given line:. Point : (,) line: 4-y=

3 III. Functions Definition: Let f and g be functions. The function given by (f g)()=f(g()) is called the composite of f with g. The domain of f g is the set of all in the domain of g such that g() is in the domain of f. Eample: Given: f()=+ and g()=- Find: f(g()), g(f()) and f(g()) To find f(g()) we must first find g(): g()=()- =4-= Since g()= we can find f(g())=f()=()+=9+=4 To find g(f()) we must first find f(): f()=()+=6+= Since f()= we can find g(f())=g()=()-=-= To find f(g()) we must put the function g() into f() equation in place of each. f(g())=f(-)=(-)+=6-+=6+ The domain of a function is the set of values for which the function is defined. The range of a function is the set of y values that a function can return. In Calculus we usually write domains and ranges in interval notation. If the domain were -< 7 then in interval notation the domain would be (-,7]. Notice that the left side has a ( because it does not include - but the right side includes 7 so we use a ]. When using interval notation we never use a [ or ] for infinity. Eample: Find the domain and range for Since we can only take the square root of positive numbers - 0 which means that. So we would say the domain is [, ). Note that we have used a [ to indicate that is included. If was not to be included we would have used (, ). The smallest y value that the function can return is 0 so the range is (0, ). Problem Set III Let f()=+ and g()=- find each of the following:. g(f()). f( + ) Find the domain and range for each function give your answer using interval notation:. h()= 4 4. f()=

4 IV. Asymptotes and Holes Given a rational function if a number causes the denominator and the numerator to be 0 then both the numerator and denominator can be factored and the common zero can be cancelled out. This means there is a hole in the function at this point. Eample: Find the holes in the following function When = is substituted into the function the denominator and numerator both are 0. Factoring and canceling: ( )( ) but ( ) this restriction is from the original function before canceling. The graph of the function f() will look identical to ( ) y ecept for the hole at =. ( ) ( ) f note the hole at = y ( ) Given a rational function if a number causes the denominator to be 0 but not the numerator to be 0 then there is a vertical asymptote at that value. Eample: Find the vertical asymptotes for the function When =- is substituted into f() then the numerator is - and the denominator is 0 therefore there is an asymptote at =. See the graphs above. Given a rational function if a number causes the numerator to be 0 but not the denominator to be 0 then the value is an -intercept for the rational function. Eample: Discuss the zeroes in the numerator and denominator When =- is substituted into the function the numerator is 0 and the denominator is -6 so the value of the function is f(-)=0 and the graph crosses the -ais at =-. Also note that for =0 the numerator is and the denominator is 0 so there is a vertical asymptote at =0. The graph is below. Problem Set IV For each function below list all holes, vertical asymptotes and -intercepts. ( )( ) ( )( ). y g ( )

5 V. Trig. Equations and Special Values You are epected to know the special values for trigonometric functions. Fill in the table below and study it. (degrees) cos sin Quadrant (degrees) (radians) (radians) cos sin Quadrant Reciprocal identities sin cos csc sec tan cot csc sin sec cos cot tan Tangent Identities sin tan cos cos cot sin Pythagorean Identities sin cos tan sec cot csc Reduction Identities sin( ) sin cos( ) cos tan( ) tan We use these special values and identities to solve equations involving trig functions. Eample: Find all solutions to sin sin sin sin sin sin 0 ( sin )(sin ) 0 ( sin ) 0 and (sin ) 0 sin and sin k 6 and k k 6 Original Problem Get one side equal to 0. Factor Set each factor equal to 0 Problem Set V Find all solutions to the equations. You should not need a calculator.. 4cos 4cos. sin sin 0. sin cos 4. sin(cos ) Get the trig function by itself Solve for (these are special values)

6 VI. Eponents A fractional eponent means you are taking a root. For eample / is the same as. Eample: Write without fractional eponent: / y y Notice that the root is the bottom number in the fraction and the power is the top number in the fraction. Negative eponents mean that you need to take the reciprocal. For eample y Eample: Write with positive eponents: 4 y 4 means ( ) and means ( ) / Eample: Write with positive eponents and without fractional eponents: / ( ) When factoring, always factor out the lowest eponent for each term. ( ). Eample: y 6 The lowest eponent for is - so can be factored from each term. Leaving y ( ) for the eponent for the 6 term we take - (-) and get. For the term we take --(-) and get as our new eponent. When dividing two terms with the same base, we subtract the eponents (numerator eponent- denominator eponent). If the difference is negative then the term goes in the denominator. If the difference is positive then the term goes in the numerator.. Notice that Eample: Simplify () 8 First you must distribute the eponent. the eponents. Since -8 results in - we know that we will have in the denominator Then since we have two terms with as the base we can subtract 8 Eample: Simplify ( ) First we must factor both the numerator and denominator. ( ) ( )( ). Then we can see that we have the term (-) in both the numerator and denominator. Subtracting eponents we get -= so the term will go in the numerator with as it s eponent. ( ) ( ) ( ) f. Eample: Factor and simplify 4( ) / ( ) /

7 The common terms are and (-). The lowest eponent for is. The lowest eponent for (-) is -/. So factor out / ( ) ( )./ f [4( ). This will simplify to ( ) / [4 ]. Leaving ( ) and obtain ] ( ) a final solution of. Problem Set VI Write without fractional eponents.. y /.. (6 y / / ) / Write with positive eponents: y (4 y ) ( 4 ) ( ) 7. ( ) Factor then simplify: / ( ) ( ) 0. 6( ) 4( ) Simplify: (4 ) ( )( ) y 4 ( ) ( ) f 6 ) 4 ( y 8 0

8 You will need to use your own paper to work the problems in this packet. You will turn in ALL of your work and the attached completed answer sheet. AP CALC Summer Packet Name Problem Set I ) -int y-int ) -int y-int ) -int y-int 4) -int y-int ) -int y-int Problem Set II ) y = ) y = ) y = 4) y = ) = 6) y = 7) y = 8) m = b = 9-) (put on the same graph) ) parallel, y = and perpendicular, y = Problem Set III ) f(g())= _ ) f( + ) = ) domain range 4 domain range

9 Problem Set IV ) holes vertical asymptotes -int ) holes vertical asymptotes -int ) holes vertical asymptotes -int 4) holes vertical asymptotes -int Problem Set V ) = _ ) = _ ) = _ 4) = _ Problem Set VI ) y = ) f () = ) y = 4) f () = ) y = 6) y = 7) f () = 8) f () = 9) f () = 0) f () = ) f () = ) y = ) f () = 4) y =