1. Math 143 Final Review Spring 2007

2. perpendicular to the given line 4x – 5y = 6 goes through (-2, 3)1.
Given line
Line in question
perpendicular to the given line
4x – 5y = 6
goes through (-2, 3)
-5y = -4x + 6
-5 4
m =
Equation:
4 5
6 5
y = x – -5
y – 3 x + 2 = 4
4 5
m =
4y – 12 = -5x – 10
4y = -5x + 2
-5 4
1 2
Slope-intercept form
y = x +
5x + 4y = 2
Standard form

3. 2. Given f(x). Draw g(x) = -f(-x + 1) + 3
1. Hor shifts
2. Flips
3. Vert shifts 3 -3 3
Shift one unit left
Flip over the y-axis
Flip over the x-axis
Shift three units up -3

4. x – 2 x + 3
f(x) = x2 – 2x g(x) = 3x h(x) =
3. (f o g)(x) =
(3x + 1)2 – 2(3x + 1)
= 9x2 + 6x + 1 – 6x – 2
= 9x2 – 1 4. g 2 2 6 = =
+ 1
x – 1
x – 1
x – 1
x – 1 x – 1 6 = +
x – 1
x + 5 x – 1 =

5. x – 2 x + 3
f(x) = x2 – 2x g(x) = 3x h(x) = 2
2 3x 2 5. = =
g(x) – 1
(3x + 1) – 1
f(x + h) – f(x) h
[(x + h)2 – 2(x + h)] – [x2 – 2x] 6. = h
x2 + 2xh + h2 – 2x – 2h – x2 + 2x = h
2xh + h2 – 2h
h(2x + h – 2) = = h h =
2x + h – 2

6. x – 2 x + 3
f(x) = x2 – 2x g(x) = 3x h(x) = 7.
h-1(x) = ?
Inverse
Original function
y – 2 y + 3
x – 2 x + 3
x =
h(x) =
xy + 3x = y – 2
x – 2 x + 3
y =
xy – y = -3x – 2
y(x – 1) = -3x – 2
-3x x – 1
y =
-3x x – 1
h-1(x) =

7. x – 2 x + 3
f(x) = x2 – 2x g(x) = 3x h(x) =
(4) – 2 (4) + 3 8.
h(4) =
2 7 =

8. It passes the vertical line test
9. Is f(x) a function?
f(x) 3
Yes
Why or why not? -3 3
It passes the vertical line test -3
Does f(x) have an inverse function? No
Why or why not?
It does not pass the horizontal line test

9. 10. a. What is the domain of f(x) ? f(x)3
(-4, 5] -3 3
b What is the range of f(x) ? -3
[0, 3]
c. What is f(5) ?
f(5) = 3

10. 11. a. For what value(s) of x does f(x) = 2 ? f(x)3
When x = 3 or x = -3 -3 3 -3
b Identify any x-intercepts
(0, 0)
c. Identify any y-intercepts
(0, 0)

11. 12. a. For what values of x is the graph of f(x) increasing ? f(x)3
From x = - 4 to x = -3 -3 3
From x = 0 to x = 5 -3
b For what values of x is the graph of f(x) decreasing ?
From x = -3 to x = 0
c. For what values of x is the graph of f(x) constant ?
None

12. 13. Is the degree of f(x) even or odd? How do you know?-3 3
The degree is odd -3
The endpoint behavior is different on the right and left.
What can be said about the leading coefficient of f(x)? Why? -9
The leading coefficient is positive.
The right side of the graph rises.

13. 15. What is the minimum degree of f(x)? How do you know?3
What is the minimum degree of f(x)? How do you know? -3 3
Minimum degree = 5 -3
The graph makes four turns.
Counting multiplicities, what is the minimum number of real zeros of f(x) -9
The minimum number of real zeros is 5.
At least 2 zeros at x = -5, at least two zeros at x = -1, and at least 1 zero at x = 3.

14. -5 even multiplicity -1 even multiplicity 3 odd multiplicity
f(x) 3
State each zero of f(x) and whether its multiplicity is even or odd. -3 3 -3
Xeros at x =
even multiplicity even multiplicity odd multiplicity
Write one possible equation for f(x) -9
f(x) = (x + 5)2(x+ 1)2(x – 3)

15. 19.
2x , x > 3
Evaluate each of the following
f(x) =
Ö7 – x , x £ 3
a. f(4)
b. f(3)
c. f(- 9)
= 2(4)2 + 4
= 36 =
= 2
Ö7 – 3 =
= 4
Ö7 + 9

16. 20. List all the possible rational zeros of:
20. List all the possible rational zeros of: f(x) = 3x4 – 13x3 + 22x2 – 18x + 4
factors of p: ±4, ± 2, ± 1
factors of q: ±3, ± 1
possible rational zeros: ±4, ± 2, ± 4/3, ±1, ± 2/3, ± 1/3

17. The other two zeros must be the solutions of 3x2 – 6x + 6 = 0
Find all the zeros of: f(x) = 3x4 – 13x3 + 22x2 – 18x + 4
By seeing the graph of the function or by plugging the values from problem 20 into the function you can determine that there are zeros at 2 and at 1/3
The other two zeros must be the solutions of 3x2 – 6x + 6 = 0
1/3 2
x2 – 2x + 2 = 0
2 ± Ö4 – 4(1)(2)
x = 2
2 ± Ö - 4 = =
1 ± i 2
Zeros: 1/3, 2, 1+ i, 1 - i

18. 22.Graph the system: x2 + y2 £ 9 and y > x + 1
Border: x2 + y2 = 9
Test (0, 0)
0 + 0 £ 9
True
Border: y = x + 1
Test (0, 0)
0 > 0 + 1
False

19. 23. Match each equation to its type
(x – 3)2 + (y + 1)2 = 16
A. Linear
B. Quadratic
C. Higher order polynomial
D. Rational
E. Circle
F. Ellipse
H. Hyperbola
J. Parabola A
y = 2x – 3
x2 + 3x + 2 D
y =
x2 – 9 H
(x – 3)2 – 2(y + 1)2 = 9 C
y = 4x4 – 3x2 + 5

20. 23. Complete the square and draw the graph
9x2 + 16y2 – 18x + 64y – 71 = 0
9(x2 – 2x + __) + 16(y2 + 4y + __) = 71 + __ + __ 1 4 9 64
9(x – 1)2 + 16(y + 2)2 = 144
(x – 1) (y + 2)2
= 1

21. 23. Complete the square and draw the graphb.
16x2 –y2 + 64x – 2y + 67 = 0
16(x2 + 4x + __) – 1 (y2 + 2y + __) = __ + __ 4 1 64 -1
16(x + 2)2 – 1(y + 1)2 = -4
(y + 1) (x + 2)2
– = 1

22. 25. Evaluate each of the following.
0! = 1 1
1! =
5! =
120

23. nCr 26. Evaluate: 8 3 8! (8)(7)(6) 5! = = 3! 5! 3! 5! = 56
8 3 8!
(8)(7)(6) 5! = =
3! 5!
3! 5! = 56
On the calculator:
8 3 8
nCr 3 = 56 =

24. 27. Find the third term of (x – 3)9
9 2
(x)7
(-3)2
3rd term =
= 36(x7)(9)
= 324x7

25. Both answers work in the original problem
Solve Ö2x + 3 – Öx – 2 = 2 2 2
Ö2x = 2 – Öx – 2
2x + 3 = 4 – 4Öx – 2 + (x – 2) 2 2
x = -4Öx – 2
x2 + 2x = 16(x – 2)
x2 – 14x = 0
(x – 3)(x – 11) = 0
Both answers work in the original problem
x = 3 or x = 11

26. 1.6 Solve: Öx – 3 Öx – 10 = 0 x1/2 – 3x1/4 – 10 = 0 Let: a = x1/4
The form of this equation looks like a quadratic equation.
x1/2 – 3x1/4 – 10 = 0
Let: a = x1/4
a2 – 3a – 10 = 0
(a – 5)(a + 2) = 0
a = 5 or a = -2 so
x1/4 = 5 or
x1/4 = -2
(x1/4)4 = (5)4
(x1/4)4 = (-2)4
x = 625
x = 16
16 does not work in the original equation, but 625 does work.
x = 625

27. Find the key values of x by solving |5x – 2| = 13
1.8a Solve: |5x – 2| > 13
Find the key values of x by solving |5x – 2| = 13
5x – 2 = or 5x – 2 = -13
5x = 15
5x = -11
x = 3
x = -11/5
Now test numbers from the intervals created by these key values. Use the original problem to test. T F T 3
-11/5
Solution: x < -11/5 or x > 3

28. Find the key values of x by solving 4x2 + 7x = -3
1.8b Solve: 4x2 + 7x < -3
Find the key values of x by solving 4x2 + 7x = -3
4x2 + 7x + 3 = 0
(4x + 3)(x + 1) = 0
x = -3/4 or x = -1
Now test numbers from the intervals created by these key values. Use the original problem to test. F T F
-3/4 -1
Solution: -1 < x < -3/4

29. 2.7 a. f(x) = (1/2)x3 – 4 Inverse Original function x = (1/2)y3 – 4
y = (1/2)x3 - 4
y3 = 2x + 8 3
y = Ö2x + 8 3
f-1(x) = Ö2x + 8

30. 3.4. f(x) = 6x3 + 25x2 – 24x + 5 one of the zeros is -5 - 5 6 25 -24 5
factors of p: ±5, ± 1
factors of q: ±6, ± 3, ±2, ± 1
possible rational zeros: ±5, ± 5/2, ± 5/3, ±1, ± 5/6, ± 1/2, ± 1/3, ± 1/6
one of the zeros is -5
The other two zeros are the solutions to
- 5
6x2 – 5x + 1 = 0
(3x – 1)(2x – 1) = 0
x = 1/ x = 1/2
zeros: , 1/3, 1/2

31. The solutions for x2 – 2x + 2 = 0 are
Find all the roots of x4 – 4x3 + 16x2 – 24x + 20 = 0 given that 1 – 3i is a root.
Since 1 – 3i is a root, i is also a root
so (x – 1 + 3i) and (x – 1 – 3i) are factors of the equation
(x – 1 – 3i)(x – 1 + 3i) =
x2 – 2x + 10
See problem 21
x2 – 2x – 2
The solutions for x2 – 2x + 2 = 0 are
x2 – 2x + 10
x4 – 4x3 + 16x2 – 24x + 20 - + -
x4 – 2x3 + 10x2
– 2x3 + 6x2 – 24x + 20
x =
1 ± i +
– 2x3 + 4x2 – 20x - +
2x2 – 4x + 20
2x2 – 4x + 20
Roots: 1 – 3i, 1 + 3i, 1 – i, 1 + i

32. 3.6 Graph the function: f(x) = x + 2
Vertical asymptote: x = -2
Horizontal asymptote: y = -3
x-intercept: (0,0)
y-intercept: (0,0)
x y -6 -9 3 -1

33. 4.2 f(x) = logb x is shown on the graph
a. When x = 5,
logb 5 = 1
f-1(x)
so b = 5 5 -5 5
c. For f-1(x)
Domain: all real numbers -5
Range: y > 0

34. log 140.3
log =
» 4.5
log 3
4.4 a. Solve: 22x – = 35
22x – 1 = 32
ln 22x – 1 = ln 32
(2x – 1) ln 2 = ln 32
2x – 1 = 5
2x = 6
x = 3

35. x = -4 does not work in the original equation.
4.4 b Solve: log3 (x – 5) + log3 (x + 3) = 2
log3 (x2 – 2x – 15) = 2
x2 – 2x – 15 = 32
x2 – 2x – 24 = 0
(x – 6)(x + 4) = 0
x = 6 or x = -4
x = -4 does not work in the original equation.
Solution: x = 6

36. A blood alcohol level of 0.112 corresponds to an accident risk of 25%
R = 6e12.77x
If R = 25% accident risk
25 = 6e12.77x
25/6 = e12.77x
ln (25/6) = ln e12.77x
ln (25/6) = x
x =
A blood alcohol level of corresponds to an accident risk of 25%

37. 7.1 Write the standard form of the equation of an ellipse given
foci: (0,-3) and (0,3)
vertices: (0,-4) and (0,4)
Center: (0, 0) V
The equation must be in the form: F
x y a b2
= 1
b = distance from the center to a vertex = 4 F
c = distance from the center to a focus point
c = 3 V
c2 = b2 – a2
9 = 16 – a2
x y
Equation:
= 1
a2 = 7

38. The parabola is vertical with an equation in the form:
7.3 Find the equation of a parabola with vertex at (2, -3) and focus at (2, -5)
The parabola is vertical with an equation in the form:
(x – h)2 = 4p(y – k)
(h, k ) = (2, -3) V
p = -2 F
Equation:
(x – 2)2 = 8(y + 3)