= ( )= To find the domain, we look at the vertical asymptote(s) (where denominator equals zero) , =0

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1 Precalculus College Algebra Review for Final Name It is also a good idea to go back through your old tests and quizzes to review. 1. Find (+1) given ()= Determine () given ()=+2 and ()= (+1)=3(+1) +1 =3( +2+1)+1 = = ()= Find the domain and range of the function ()= To find the domain, we look at the vertical asymptote(s) (where denominator equals zero) (+9)( 4) 0 9, 4 To find the range, we look at the horizontal asymptote 4. Find the domain of the function ()= 5+2 lim =0 0 The denominator cannot equal zero and we cannot do the square root of a negative number, so we solve 5. Graph ()= >0 > 2

2 6. Graph ()= Determine the vertex of ()= Vertex is at =, then plug in that x-value in the equation to find the y-value

3 = 24 = = 24 6 = 4 ( 4)=3( 4) +24( 4) 1= = = 48 1= 49 ( 4, 49) 8. Determine the equation of a parabola with vertex ( 5, 1) that goes through the point (3,1) The general equation is ()=( h) +. The vertex tells us (h,) and then we can plug in the other point to solve for. 9. Find the inverse function of ()=7+2 ()=(+5) 1 1=(3+5) 1 2=(8) 2=64 = 1 32 ()= 1 32 (+5) 1 First we switch the x and y values, then we solve for y =7+2 2= = ()= Find each zero and its multiplicity then graph ()=(+3) (4+3)( 4) To find the zeros we set the y-value equal to zero and solve for x 0=(+3) (4+3)( 4) 0=(+3) ;0=4+3;0= 4 0=+3; 3=4;4=

4 3=; 3 =; ±2= 4 = 3 ( 2), 3 ( 1),2 ( 1), 2 ( 1) 4 To graph use the zeros along with odd multiplicities crossing and even multiplicities touching along with at least one test point 11. Determine all of the possible rational zeros, find each zero and its multiplicity, and then graph ()= To find the zeros we look at factors of the last term over factors of the first term ± 1,2,4,5,10,20 1 =±1,±2,±4,±5,±10,±20 To help us find what zero to attempt with the synthetic division we can look at its graph

5 This function appears to have a root at 2 with an even multiplicity: Now we have factored =( 2)( ) and we now have to solve =0, so we will continue with the synthetic division using the root of 2 again Now we have factored =( 2) ( +5) and we need to solve + 5=0 +5=0 = 5 =± 5 =± 5 =2( 2); 5( 1); 5( 1) I already showed the graph above

6 12. Find each zero and its multiplicity, the y-intercept, the vertical asymptote(s), the horizontal asymptote, and then graph ()= Zeros will occur where the y value is 0: y-intercept will occur where the x value is 0: 0= =+2 = 2 ( 2,0) = = 2 9 0, 2 9 Vertical asymptotes will occur where the y-value goes to infinity, which happens when the bottom equals zero: 9=0 =9 =±3 Horizontal asymptote will occur where the x-value goes to infinity: = lim =0

7 13. Write the equation of a polynomial with rational coefficients and the zeros 3,2,5 ()=( 3) (2 ) (2+)( 5)(+5) ()=( 3)( 2+)( 2 )( 5)(+5) 14. Determine the remainder of (4 2 +1) (+3) Thus the remainder is Determine the maximum number of turning points of ()= Since the degree is 7, the maximum number of turning points (aka local max s and local min s) is Graph then determine the domain, range, and horizontal asymptote of ()= Using a base graph of 2 you can determine the points x y -1 ½ Along with an base asymptote of =0. Using transformations of +1; 3+4 you end up with the graph

8 :R;: >4;h :=4 17. Graph then determine the domain, range, and vertical asymptote of ()=2 log (+2)+1 Using a base graph of log we can determine the following points using the inverse 3 Inverse Base x y x y -1 1/3 1/ Along with a vertical asymptote of =0. Using transformations of 2; 2+1 you end up with the graph 18. Solve the equation 7 = Solve the equation 3( +2)=60 : > 2;:R; := 2 log 7 =log 20 =log 20= ln20 ln ( +2) = =20

9 =18 = 18 = Solve the equation 2log (+4)=9 2log (+4) = log (+4)= () = Solve the equation log( 2)+log( 3)=2 log( 2)( 3)=2 log( 5+6)=2 10 =10 5+6= =0 = 5±( 5) 4(1)( 94) 2(1) 12.51, 7.51 Removing the extraneous solution we have the solution: 22. Rewrite ln = 5± = 5± using the sum or difference of logarithms with powers expressed as factors ln +ln ln =3ln+2ln 5ln

10 23. Rewrite 2log log 4+3log (2) as a single logarithm log log 4+log (2) =log log 4+log 8 8 =log 4 =log If Bob deposits $300 at 3.7% interest compounded daily, how much will he have after two years? =1+ = =$ If Sally invests $400 at 7% interest compounded continuously, how long will it take the account to reach $700? = 700= = =. ln 7 4 =ln. ln 7 4 =0.07 ln = Ann bought a $250,000 house with a 30-year mortgage at 5.5% APR. How much will her monthly payment be?

11 = ,000= ,000= =$ Sue is investing $500 per month into a retirement fund that will earn 9% APR. What is the value of her account after 35 years? = 1+ 1 = =$1,470, Find the vertex, focus, directrix, focal width, and then graph = (+1) 1 :( 1, 1) Since it opens up, we will be adding to the y-value of the vertex to get the focus 1 4 = = 1 = 2 = :( 1, 1+2)=( 1,1) := 1 2 = 3 h: 1 =1 = 8 =8 1 8

12 29. Find the vertex, focus, directrix, focal width, and then graph =2( 2) +5 :(5,2) Since it opens right, we will be adding to the x-value of the vertex to get the focus 1 4 = =1 8 =1 8 :5+ 1 8,2=51 8,2 :=5 1 8 =47 8 h: 1 =1 2 = Put the equation into graphing form and state the coordinates of the vertex = =3 4( 2+) 5 =3 4( 1) 9=3

13 4 3 ( 1) 3= :( 3,1) 31. Determine the equation of the parabola given a vertex of (2,3) and focus (4,3) When you sketch a graph you can see that it opens to the right, which means that it will be = with a positive a value The coordinates of the vertex give us h and k h=2;=3 The distance between the vertex and the focus is 2, so we can set that equal to and solve for a 2= 1 4 2= 1 4 ; 2= 1 4 2= 1 4 ; 2=1 4 =± 1 8 We know a has to be positive, so the final equation is = 1 8 ( 3) Determine the equation of the parabola given a focus of (6, 2) and directrix =0

14 When you sketch a graph you can see that it opens down, which means that it will be = and a will be negative The vertex has to be exactly halfway between the directrix and focus, so that gives us h and k h=6;= 1 The distance between the vertex and focus is 1, so we set that equal to to solve for a 1= 1 4 1= 1 4 ; 1= 1 4 =± 1 4 We know a has to be negative, so the final equation is = 1 4 ( 6) Find the center, foci, and vertices, then graph () + () =1 :(1, 4) :(1, 4±8) (1,4) &(1, 12) = = = 39 =39 = 39 :1, 4± 39

15 34. Find the center, foci, and vertices, then graph () () =1 :(3, 2) :(3, 2±3) (3,1) &(3, 5) = + =9+16 =25 =5 :(3, 2±5) (3,3) &(3, 7)

16 35. Determine the equation of the ellipse or hyperbola given foci ( 3,4)&( 3,2) and vertices ( 3,5)&( 3,1) Sketching a graph of the information we can tell it is an ellipse because the foci are inside the vertices. We can also tell that the major axis is vertical, so > The center has to be directly between the foci and vertices, so that tells us h and k h= 3;=3 The distance from the center to the vertices tells us b =2 The distance from the center to the foci tells us f, so we can use f and b to solve for a =1 = 1 = 2 1= 4; 1= 4 5= ;3= = 5, 3 We know that >, so a must be 3 and we can write the equation (+3) 3 + ( 3) Determine the equation of the ellipse or hyperbola given center (2,7); focus (5,7); and vertex (4,7) Sketching a graph of the information we can tell it is a hyperbola because the foci are outside the vertices. We can also tell that the transverse axis is horizontal, so the x part is positive =1

17 The center tells us h and k h=2;=7 The distance from the center to the vertices tells us a =2 The distance from the center to the foci tells us f, so we can use f and a to find b ( 2) 4 =3 = + 3 =2 + 9=4+ 5= = 5 ( ) Put the equation in graphing form and state the center for =0 = =19 4( +10+) 9( +8+) ( )=19 4(+5) 9(+4) 44=19 4(+5) 9(+4) = 36

18 4(+5) 36 (+5) 9 (+4) 4 9(+4) 36 (+4) 4 (+5) 9 :( 5, 4) = =1 =1