A.P. Calculus Summer Assignment This assignment is due the first day of class at the beginning of the class. It will be graded and counts as your first test grade. This packet contains eight sections and an answer sheet. Each section contains necessary definitions, eamples, solutions and a problem set. Read the definitions, eamples and solutions in order to solve all problems in each problem set. Summer Assignment Checklist: o Start assignment early giving yourself ample time to complete it. o If you are unable to solve any problem on your own then use math reference books, websites, and/or your notes to find solutions. o Write all answers on answer sheet. o Use pencil to complete assignment. o Since you must show work in order to receive full credit, attach separate sheets of paper. Only submitting answers will result in no credit! o Complete all problems in handout. o Complete prior to first day of school. o You may contact me by email at Charlene.Jones@pgcps.org if you have questions. Note: Assignments NOT received on first day of class will not be accepted. Students not submitting summer assignment on first day of class will receive a grade of zero.
I. Symmetry: -ais If a graph is symmetric to the -ais for every point (,y) on the graph the point (,-y) is also on the graph. To test for symmetry about the -ais substitute in y for y into the equation if this yields an equivalent equation then it is symmetric. y-ais If a graph is symmetric to the y-ais for every point (,y) on the graph the point (-,y) is also on the graph. To test for symmetry about the y-ais substitute in for into the equation if this yields an equivalent equation then it is symmetric. Origin If a graph is symmetric to the origin for every point (,y) on the graph the point (-,-y) is also on the graph. To test for symmetry about the origin substitute in for into the equation and substitute in y for y into the equation if this yields an equivalent equation then it is symmetric. Eample: Test for symmetry with respect to each ais and the origin. Given equation: y 4 0 Y-ais: (change all to ) y 4 ( ) 0 y 4 0 since there is no way to make this look like the original it is NOT symmetric to the Y-ais X-ais (change all y to y): ( y) 4 0 y 4 0 since there is no way to make this look like the original it is NOT symmetric to the X-ais origin: (change all to and change all y to y ) ( y) 4 ( ) 0 y 4 0 since this does look like the original it is symmetric to the origin. The figure to the right shows the graph of y 4 0. It is symmetric only to the origin. Problem Set I Test for symmetry with respect to each ais and the origin. Be sure to show your work.. y 4 +. y +. + y 4.. y Page
II. Intercepts The -intercept is where the graph crosses the -ais. You can find the -intercept by setting y0. 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 set y0 4 + ± ( + ) take square root of both sides ( + ) or ( + ) Write as equations or Subtract from both sides ( ) add 4 to both sides y-intercept y (0 + ) 4 set 0 Add 0+ y 4 y 9 4 Square y subtract Problem Set II Find the intercepts for each of the following.. y 6 + 4. y 4 0. y + 4. y 6. y 6 III. Lines The slope intercept form of a line is ym+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 ( y, ) is a point on the line. If two lines are parallel then they have the same slope. If two 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) ( ) + b Using the slope-intercept form with m and point (-,) Page
Multiply -and - + b b Subtract from both sides b Get a common denominator of b Combine like terms This is the equation of the line parallel to the given line that contains (-,) y + Part b (using slope from eample above) ( ) + b Using the slope-intercept form with m and point (-,) 6 + b Multiply and - 6 b Subtract -6 from both sides 7 b subtract y + 7 This is the equation of the line perpendicular to the given line that contains (-,) 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 the point (,-) and has slope m -. Since we are given a point and slope it is easiest to use the point slope form of a line. y ( ) Substitute into point slope form for (, y) and m y + + Minus a negative makes + and distribute - y Subtract from both sides 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 (4) + b Using the slope-intercept form with m and point (4,) 8 Multiply and 4 + b 8 b Subtract 8 from both sides 8 b Get a common denominator of 7 b Combine like terms We know the equation of the line is 7 y in slope intercept form. Page 4
Remember that when we graph lines that we plot the y-intercept first and then from that point we graph the slope using rise over run. Eample: Graph the following equation: y + First you must get the equation in slope-intercept form: y y The slope / and the y-intercept - Plot the point (0,-). From the first point go up and over to the right to get a second point.. Now connect the two points to get the line. Problem Set III Find the equation of a line in slope intercept form:. contains (,-4) and (,). contains, and, 4 6. contains (,) and m0 4. contains (,7) and m-. contains (-,4) and m is undefined 6. contains (-,-) and m 7. contains (-,4) and m 8. contains (0,0) and (,6) 9. contains (-,6) and (,) 0. contains (,-) and (,-). -intercept (,0) and y-intercept (0,). -intercept,0 and y-intercept (0,-). contains (-,4) and -intercept (a,0) and y- intercept (0,a) and a 0 Find the slope and y-intercept of the line: 4. +y 0. 6-y 6. 4 7. y- Sketch a graph of the equation: 8. y- 9. 4 0. y-+. +y+60 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. Point: (-,) : line: +4y7 Page
IV. Functions In order for a graph to represent a function it must be true that for every value in the domain there is eactly one y value. To test to see if an equation is a function we can graph it and then do the vertical line test. Eample: Is y a function? The graph is below: When a vertical line is drawn it will cross the graph more than one time so it is NOT a function. 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, ). Hint: In general, set the epression under the square equal to zero and solve as we did in the preceding eample. If there is an epression in the denominator set that equal to zero and solve. Problem Set IV Sketch the graph of the equation and then use the vertical line test to determine whether it is a function of.. y. y. 9 y Let f()+ and g()- find each of the following: 4. f(g(0)) 6. g(f()). g(g()) 7. g(f()) 8. f(g()) Find the domain and range for each function give your answer using interval notation: 9. h(). h() +4 0. h() 4. f() +, <. f() +, Page 6
V. 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. ( + ) 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. Page 7
Eample: Find the holes, vertical asymptotes and -intercepts for the given function: f ) + 6 ( First we must factor to find all the zeroes for both the numerator and denominator: ( ) ( + ) Numerator has zeroes 0 and Denominator has zeroes 0 and -. 0 is a hole - is a vertical asymptote is a -intercept Problem Set V For each function below list all holes, vertical asymptotes and -intercepts. ( )( + ) ( )( + ).. 4. y + + 8 9 + 4 g ( ) + + VI. 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 ( ϑ ) Radians ϑ Quadrant Sine ϑ Cosine ϑ 0 0 4 60 90 0 0 80 0 40 70 00 0 Page 8
You should study the following trig identities and memorize them before school starts: Reciprocal identities sin cos csc csc sin sec sec cos tan cot cot tan Tangent Identities sin tan cos cos cot sin Pythagorean Identities sin + cos tan + sec cot + csc Double angle Identities sin sin cos cos cos sin cos sin 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 π + πk 6 π and + πk π + πk 6 Original Problem Get one side equal to 0. Factor Set each factor equal to 0 Problem Set VI Get the trig function by itself Solve for (these are special values) Complete chart on answer sheet. Find all solutions to the equations. You should not need a calculator.. 4cos 4cos. sin + sin + 0. sin sin 0 4. sin cos. sin + sin 6. cos + cos (hint: use double angle identity) 7. sin(cos ) 8. sin sin 0 Page 9
VII. Absolute value and piecewise functions In order to remove the absolute value sign from a function you must:. Find the zeroes of the epression inside of the absolute value.. Make sign chart of the epression inside the absolute value.. Rewrite the equation without the absolute value as a piecewise function. For each interval where the epression is positive we can write that interval by just dropping the absolute value. For each interval that is negative we must take the opposite sign. Eample: Rewrite the following equation without using absolute value. + 4 +40 Find where the epression is 0-4 Subtract 4-4/ Divide by X- Simplify - + - Put in any value less than - into +4 and you get a negative. Put in any value more than - into +4 and you get a positive. 4 + 4 < Write as a piecewise function. Be sure to change the signs of each term for any part of the graph that was negative on the sign chart. Eample: Rewrite the following equation without using absolute value. + +- 0 Find where the epression is 0 (-)(+)0 factor -0 or +0 Set each factor equal to 0 X/ or - Solve each equation + - + Put in any value less than - into (-)(+) and you get a positive number. - + + / < and > / / Put in any value more than - and less than ½ into (-)(+) and you get a negative number. Put in any value more than / into (-)(+) and you get a positive number. Write as a piecewise function. Be sure to change the signs of each term for any part of the graph that was negative on the sign chart. Page 0
Eample: Rewrite the following equation without using absolute value. -90 9 + 9 Add 9 Divide by Find where the epression is 0 (For the part in the absolute value only.) + 9 + 9 + + 7 - + < < Put in any value less than into -9 and you get a negative. Put in any value more than into -9 and you get a positive. Write as a piecewise function. Be sure to change the signs of each term that is inside the absolute value for any part of the graph that was negative on the sign chart. Simplify. Problem Set VII Rewrite the following equation without using absolute value. Be sure to show your work, including a sign chart:. +. ( + )( 4). 7 + 4. +. + 6 VIII. Eponents A fractional eponent means you are taking a root. For eample / is the same as. Eample: Write without fractional eponent: / y fraction. y Notice that the root is the bottom number in the fraction and the power is the top number in the Negative eponents mean that you need to take the reciprocal. For eample. means and means y Eample: Write with positive eponents: 4 y 4 Eample: Write with positive eponents and without fractional eponents: f ( + ) ( ) / ( ) / ( ) ( + ) Page
When factoring, always factor out the lowest eponent for each term. Eample: y + 6 can be factored from each term. Leaving The lowest eponent for is - so. Notice that for the eponent for the 6 term we take - (-) and get. For the y ( + ) 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. Eample: Simplify () 8 First you must distribute the eponent. we can subtract the eponents. Since -8 results in - we know that we will have in the denominator. 8. 8. Then since we have two terms with as the base 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( ) + ( ) / The common terms are and (-). The lowest eponent for is. The lowest eponent for (-) is -/. So factor out / ( ) and obtain ( )./ [4( ) + ] ( ) / [4 + ]. Leaving a final solution of ( ). This will simplify to. Page
Problem Set VIII Write without fractional eponents.. / y. 4 (6 ) /. y / / 4 7 Write with positive eponents: 4.. 6. y (4 y ) ( 4 ) f ( ) ( ) 7. ( + ) Factor then simplify: 8. 4 + 8 9. 0. / ( ) / + ( ) f ( ) 6( ) 4( ) Simplify:... 4. (4 ) ( + )( ) y 4 ( ) ( ) 6 8 + 4 + y 0 + Page