FUNCTIONS To evaluate a function for a given value, simply plug the value into the function for x. Recall: (f 0 g ) (x) = f(g(x)) OR f[g(x)] read 'Jofg of x" Means to plug the inside function (in this case g(x) ) in for x in the outside function (in this case, f(x)). Example: Given f(x) ==x+ 1 and g(x) = x -4 find f(g(x)). f(g(x)) = f(x-4) = (x-4) +1 =(x -8x+ 16)+1 =x -16x+3+1 f(g(x)) = X-16x + 33 Let f(x) = x+l and g(x) = X-1. Find each. L f()=. g(-3)= _ 3. f(t+l)= _ 4. f[g(-)]= _ 5. g[f(m+)]=_~_ 6. [f(x) t-g(x) = Let f(x) = sin(x) Find each exactly. 7. f(;)= 8. f(;)= Letj(x)=x, g(x)=x+5, and h(x)=x-1. Find each. 9. h[f(-)]= to. f[g(x-l)]= 4
INTERCEPTS OF A GRAPH To find the x-intercepts, To find the y-intercepts, let y = 0 in your equation and solve. let x = 0 in your equation and solve. y Example: Given the function y = x - x - 3, fmd all intercepts. x-int. (Let y = 0) o =x -x-3 0=(x-3)(x+l) x=-l or x=3 x-intercepts (-1,0) and (3,0) y-int. (Letx = 0) y = 0 -(0)-3 y=-3 y-intercept (0,-3) Find the x and y intercepts for each. 1. y=x-5 13. y=x+x- 14. 15. 5
POINTS OF INTERSECTION Use substitution or elimination method to solve the system of equations. Remember: You are finding a POINT OF INTERSECTION so your answer is an ordered pair. CALCULATOR TIP Remember you can use your calculator to verify your answers below. Graph the two lines then go to CALC (lld Trace) and hit INTERSECT. Example: Find all points of intersection of x - Y = 3 x-y=l ELIMINATION METHOD Subtract to eliminate y x -x= x -x- =0 (x-)(x+1)=0 x= or x=-l Plug in x = and x = -lto find y Points of Intersection: (,1) and (-1,-) SUBSTITUTION METHOD Solve one equation for one variable. y=x -3 y=x-l Therefore by substitution x - 3 = x -1 x-x-=0 From here it is the same as the other example Find the point(s) of intersection of the graphs for the given equations. 16. x+y=8 4x-y=7 17. x +y=6 x+y=4 18. X=3-y y=x-1 6
DOMAIN AND RANGE Domain - All x values for which a function is defmed (input values) Range - Possible y or Output values EXAMPLE I EXAMPLE Find the domain and range of f(x) Write answers in interval notation. =.J4-x1 Q;) ";""4 ~\l d' 1:tA...,. ~ 9"'). ""T'\-c..~i \,,~ft.-i- 6<-ln~""'W. fo~.~~ "r....u\ves\"vn.1~-lwa..hori~ Cl..)(.h. 'The+vrl'W!~"r \ef+1npj,,,,,q\yt <U.\oocnea.wn+-. o.l>hon-fla. 9. l"'" i -3. ~ ~ur_s,"'1\9ht ~\'(>.\~S M«.'i~ \~ """. Ch-"'L~n ;,. "6. $<> ~ \.,., \:0. C-~'b::\~~+ Is 0.\\ ~o.\s~ -?~"$. -n-e. Y'Q~se r-~..:lb ~ sat of- O>l\-~'l't>.\"'S ~ ~ ~,",~+\.w..~'a\-.je> 't'''*'' Q.\.~.~. ve.v+\c.t>.,\ o.."i.ts. ~ \. ~"I'-~,Jr 'A\._ "*~~*~\ S -. "1h~ nisn.$t it>:l..5b~ ~ hl> [-'., I],Q.nr't. l~ ~~ -,. +tl ~. DOMAIN For f(x) to be defined 4- x ~ O. This is true when -::;; x ::;; Domain: [-,] RANGE The solution to a square root must always be positive thus f (x) must be greater than or equal to O. Range: [0,00) Find the domain and range of each function. Write your answer in INTERVAL notation. 0. f(x)=--jx+3 1. f(x)=3sinx. f(x) = x-i 7
INVERSES To find the inverse of a function, simply switch the x and the y and solve for the new "y" value. Recall f-i ( x) is defined as the inverse of f ( x ) Example 1: f{x)=:vx+l Rewrite f{x) as y y = :Vx+ I x = ~y+l Switch x and y Solve for your new y ( X)3 = (~y + 1r Cube both sides x3 =y+l y-x- 3-1 f-l{x)=x3-1 Simplify Solve for y Rewrite in inverse notation Find the inverse for each function. 3. f(x) = x+ 1 x 4. f(x)=3 5 5. g(x) = x- 6. Y =.J4 - x + 1 7. Tfthe graph of f(x) has the point (, 7) then what is one point that will be on the graph of f-1 (x)? 8. Explain how the graphs of f(x) and f-i (x) compare. 8
EQUATION OF A LINE Slope intercept form: y == mx +b Point-slope form: y - Yl = m( x - Xl) * LEARN! We will use this formula frequently! Vertical line: x = c (slope is undefined) Horizontalline: y = c (slope is 0) Example: Write a linear equation that has a slope of 1; and passes through the point (, -6) Slope intercept form y =.!.x+b Plug in ~ for 17'1-6 =..!.()+b Plug in the given ordered b=-7 I y=-x-7 Solve for b I Point-slope y+6=-(x-) 1 y=-x-7 form Plug in all variables Solve for y 9. Determine the equation of a line passing through the point (5, -3) with an undefined slope. 30. Determine the equation of a line passing through the point (-4, ) with a slope of O. 31. Use point-slope form to find the equation of the line passing through the point (0,5) with a slope of /3. 3. Use point-slope form to find a line passing through the point (, 8) and parallel to the line y ::;::: ~ x-i. 6 33. Use point-slope form to fmd a line petpendicularto y =-x+9 passing through the point (4, 7). 34. Find the equation of a line passing through the points (-3, 6) and (1, ). 35. Find the equation of a line with an x-intercept (, 0) and a y-intercept (0, 3) 9
UNIT CIRCLE Yon can determine the sine or the cosine of any standard angle on the unit circle. The x-coordinate of the circle is the cosine and the y-coordinate is the sine of the angle. Recall tangent is defmed as sin/cos or the slope of the line.. 1f I sm-= Examples: 1f 1f cos-=o tan-=und *you must have these memorized OR know how to calculate their values without the use of a calculator. 36. a.) sin1f b.) cos 31f c.) sin ( -;) I{ e.) cos 4 f.) cos(-i{) g)cos':: 3 h) sin SI{ 6 1f i) cos 3 J') tan- 1f 4 k) tan1f I{ 1) tan3" 4ff m)cos- 3 n) sin 1Iff 6 0) tan 71f 4 10
TRIGONOMETRIC EQUATIONS Solv~each of the equations for 0 ~ x < 1C 37. sinx=-" 1 38. cosx =.J3 39.4sinzx=3 **Recall sinx = (sinx) **Recall if x = 5 then x =±5 40. cos x-l-cosx=o *Factor TRANSFORMATION OF FUNCTIONS h(x) = f(x) +c h(x) = f(x)-c h(x)=-f(x) Vertical shift c units up Vertical shift c units down Reflection over the x-axis h(x) = f(x-c) h(x) = f(x+c) Horizontal shift c units right Horizontal shift c units left 41. Given f(x) = x and g(x) = (x - 3) +1. How the does the graph of g(x) differ from f(x)? 4. Write an equation for the function that has the shape of f(x) = x3 but moved six units to the left and reflected over the x-axis. 43. If the ordered pair (, 4) is on the graph of f(x), fwd one ordered pair that will be on the following functions: a) f(x)-3 b) f(x-3) c) f(x) d) f(x-)+1 e) -f(x) 11
VERTICAL ASYMPTOTES Determine the vertical asymptotes for the function. Set the denominator equal to zero to find the x-value for which the function is undefined. That will be the vertical asymptote given the numerator does not equal 0 also (Remember this is called removable discontinuity). Write a vertical asymptotes as a line in the form x = 118 'i 1 Example: Find the vertical asymptote of y = _1_ x- Since when x = the function is in the form 1/0 then the vertical line x = is a vertical asymptote of the function. -9-6 -4 y 6 :1 Y=x- 6 I 44. f(x)=- x x 45. f(x) = x -4 +x 46. f(x) = x(i-x) 4-x 47. f(x) = x-16 x-i 48. f(x) = x +x- 49. f(x) = 5x+ 0 x-16 1
HORIZONTAL ASYMPTOTES Determine the horizontal asymptotes using the three cases below. Case I. Degree of the numerator is less than the degree ofthe denominator. The asymptote is y = O. Example: y == _1_ (As x becomes very large or very negative the value of this function will x-i approach 0). Thus there is a horizontal asymptote at y == 0. Case II. Degree of the numerator is the same as the degree of the denominator. The asymptote is the ratio of the lead coefficients. Exmaple: y = X : x -1 (As x becomes very large or very negative the value of this function will 3x +4 approach /3). Thus there is a horizontal asymptote at y == ~. 3 Case ill. Degree of the numerator is greater than the degree of the denominator. There is no horizontal asymptote. The function increases without bound. (lfthe degree of the numerator is exactly 1 more than the degree ofthe denominator, then there exists a slant asymptote, which is determined by long division.) Example: y == X +x -1 (As x becomes very large the value of the function will continue to increase 3x-3 and as x becomes very negative the value of the function will also become more negative). Determine an Horizontal Asymptotes. 50. f(x) = x 3 X -x+ +x-7 1 51. f(x) == 5x3 -x +8 4x-3x3 +5 53. f(x) = (x-5) x-x *This is very important in the use of limits. * 13