Math 143 Final Review

Spring 2007

1.

4x – 5y = 6

Given line Line in question

goes through (-2, 3)

-5y = -4x + 6

y = x – 6 5

4 5

perpendicular to the given line

m = 4 5

m = -5 4

Equation: -54 = y – 3

x + 2

4y – 12 = -5x – 10

4y = -5x + 2

y = x + 1 2

-5 4

Slope-intercept form

5x + 4y = 2 Standard form

3

-3 3

-3

2. Given f(x). Draw g(x) = -f(-x + 1) + 3

1. Hor shifts2. Flips3. Vert shifts

•Shift one unit left•Flip over the y-axis•Flip over the x-axis•Shift three units up

f(x) = x2 – 2x g(x) = 3x + 1 h(x) = x – 2 x + 3

3. (f o g)(x) = (3x + 1)2 – 2(3x + 1)

= 9x2 + 6x + 1 – 6x – 2

= 9x2 – 1

4. g 2x – 1 = 3 + 12

x – 1 6

x – 1 = + 1

6x – 1 = +

x – 1 x – 1

=x + 5 x – 1

f(x) = x2 – 2x g(x) = 3x + 1 h(x) = x – 2 x + 3

5.

= =

6.

= 2x + h – 2

2g(x) – 1

2(3x + 1) – 1

2 3x

f(x + h) – f(x) h =

[(x + h)2 – 2(x + h)] – [x2 – 2x]h

=

x2 + 2xh + h2 – 2x – 2h – x2 + 2xh

=

2xh + h2 – 2h h =

h(2x + h – 2) h

f(x) = x2 – 2x g(x) = 3x + 1 h(x) = x – 2 x + 3

7.

h-1(x) = ?

Original function Inverse

h(x) = x – 2 x + 3

y = x – 2 x + 3

x = y – 2 y + 3 xy + 3x = y – 2

xy – y = -3x – 2

y(x – 1) = -3x – 2

y = -3x - 2 x – 1

h-1(x) = -3x - 2 x – 1

f(x) = x2 – 2x g(x) = 3x + 1 h(x) = x – 2 x + 3

8.

h(4) =(4) – 2 (4) + 3

= 2 7

3

3

-3

-3

9. Is f(x) a function?

Yes

Why or why not?

It passes the vertical line test

Does f(x) have an inverse function?

No

Why or why not?

It does not pass the horizontal line test

f(x)

3

3

-3

-3

10. a. What is the domain of f(x) ?

(-4, 5]

b. What is the range of f(x) ?

c. What is f(5) ?

f(5) = 3

f(x)

[0, 3]

3

3

-3

-3

11. a. For what value(s) of x does f(x) = 2 ?

When x = 3 or x = -3

b. Identify any x-intercepts

c. Identify any y-intercepts

(0, 0)

f(x)

(0, 0)

3

3

-3

-3

12. a. For what values of x is the graph of f(x) increasing ?

From x = - 4 to x = -3

b. For what values of x is the graph of f(x) decreasing ?

c. For what values of x is the graph of f(x) constant ?

None

f(x)

From x = -3 to x = 0

From x = 0 to x = 5

3

3

-3

-3

13. Is the degree of f(x) even or odd? How do you know? The degree is odd

14. What can be said about the leading coefficient of f(x)? Why?

f(x)

The leading coefficient is positive.

The endpoint behavior is different on the right and left.

-9

The right side of the graph rises.

3

3

-3

-3

15. What is the minimum degree of f(x)? How do you know? Minimum degree = 5

16. Counting multiplicities, what is the minimum number of real zeros of f(x)

f(x)

The minimum number of real zeros is 5.

The graph makes four turns.

-9

At least 2 zeros at x = -5, at least two zeros at x = -1, and at least 1 zero at x = 3.

3

3

-3

-3

17. State each zero of f(x) and whether its

multiplicity is even or odd.

Xeros at x =

18. Write one possible equation for f(x)

f(x)

f(x) = (x + 5)2(x+ 1)2(x – 3)

-5 even multiplicity -1 even multiplicity 3 odd multiplicity

-9

19. f(x) =

2x2 + 4 , x > 3

7 – x , x 3

Evaluate each of the following

a. f(4)

b. f(3)

c. f(- 9)

= 2(4)2 + 4 = 36

= 7 – 3 = 2

= 7 + 9 = 4

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

21. 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

1/3 3 -13 22 -18 4

3 -12 18 -12 01 -4 6 -4

2

3 -6 6 06 -1 12

The other two zeros must be the solutions of

3x2 – 6x + 6 = 0

x2 – 2x + 2 = 0

x = 2 ± 4 – 4(1)(2)

2

= 2 ± - 4

2 = 1 ± i

Zeros: 1/3, 2, 1+ i, 1 - i

Border: x2 + y2 = 9

Border: y = x + 1

22.Graph the system: x2 + y2 9 and y > x + 1

Test (0, 0)0 + 0 9 True

Test (0, 0)0 > 0 + 1 False

23. Match each equation to its type

(x – 3)2 + (y + 1)2 = 16

y = 2x – 3

y =

x2 + 3x + 2x2 – 9

(x – 3)2 – 2(y + 1)2 = 9

y = 4x4 – 3x2 + 5

A. Linear

B. Quadratic

C. Higher order polynomial

D. Rational

E. Circle

F. Ellipse

H. Hyperbola

J. Parabola

E

A

D

H

C

9x2 + 16y2 – 18x + 64y – 71 = 0

9(x – 1)2 + 16(y + 2)2 = 144

23. Complete the square and draw the graph

9(x2 – 2x + __) + 16(y2 + 4y + __) = 71 + __ + __

(x – 1)2 (y + 2)2

16 9 + = 1

1 94 64

a.

16x2 –y2 + 64x – 2y + 67 = 0

16(x + 2)2 – 1(y + 1)2 = -4

23. Complete the square and draw the graph

16(x2 + 4x + __) – 1 (y2 + 2y + __) = -67 + __ + __

(y + 1)2 (x + 2)2

4 9 – = 1

4 641 -1

b.

25. Evaluate each of the following.

0! =

1! =

5! =

1

1

120

26. Evaluate:

8 3

=8!

3! 5!=

(8)(7)(6) 5!

3! 5!

= 56

On the calculator:

8 3

= 8nCr

3 = 56

27. Find the third term of (x – 3)9

3rd term = 9 2

(x)7 (-3)2

= 36(x7)(9)

= 324x7

1.5 Solve 2x + 3 – x – 2 = 2

2x + 3 = 2 – x – 22 2

2x + 3 = 4 – 4x – 2 + (x – 2)

x + 1 = -4x – 2 2 2

x2 + 2x + 1 = 16(x – 2)

x2 – 14x + 33 = 0

(x – 3)(x – 11) = 0

x = 3 or x = 11 Both answers

work in the original problem

1.6 Solve: x – 3 x – 10 = 0 4

x1/2 – 3x1/4 – 10 = 0

The form of this equation looks like a quadratic

equation.

Let: a = x1/4a2 – 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

x = 625

(x1/4)4 = (-2)4

x = 16

16 does not work in the original equation, but 625

does work.

x = 625

1.8a. Solve: |5x – 2| > 13

Find the key values of x by solving |5x – 2| = 13

5x – 2 = 13 or 5x – 2 = -13

5x = 15

x = 3

5x = -11

x = -11/5

Now test numbers from the intervals created by these key values. Use the original problem to test.

-11/5 3

T F T

Solution: x < -11/5 or x > 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.

-1 -3/4

F T F

Solution: -1 < x < -3/4

2.7 a. f(x) = (1/2)x3 – 4

Original function Inverse

f(x) = (1/2)x3 - 4

y = (1/2)x3 - 4

x = (1/2)y3 – 4

x + 4 = (1/2)y3

y3 = 2x + 8

y = 2x + 8 3

f-1(x) = 2x + 8 3

3.4. f(x) = 6x3 + 25x2 – 24x + 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

6 25 -24 5- 5

6 -5 1 0-30 25 - 5

The other two zeros are the solutions to

6x2 – 5x + 1 = 0

(3x – 1)(2x – 1) = 0

x = 1/3 x = 1/2

zeros: - 5, 1/3, 1/2

3.5 Find all the roots of x4 – 4x3 + 16x2 – 24x + 20 = 0 given that 1 – 3i is a root.

• Since 1 – 3i is a root, 1 + 3i 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

x2 – 2x + 10 x4 – 4x3 + 16x2 – 24x + 20

x2 – 2x – 2

x4 – 2x3 + 10x2 -+-

– 2x3 + 6x2 – 24x + 20– 2x3 + 4x2 – 20x ++ -

2x2 – 4x + 20 2x2 – 4x + 20

The solutions for x2 – 2x + 2 = 0

are

1 ± i x =

See problem 21

Roots: 1 – 3i, 1 + 3i, 1 – i, 1 + i

3.6 Graph the function: f(x) = -3xx + 2

Vertical asymptote: x = -2

x-intercept: (0,0)y-intercept: (0,0)

Horizontal asymptote: y = -3

x y

-4 -3 -1 1

-6-93

-1

4.2 f(x) = logb x is shown on the graph

5

5

-5

-5

a. When x = 5,logb 5 = 1so b = 5

f-1(x)

c. For f-1(x)

Domain: all real numbers

Range: y > 0

4.3 log3 140.3 =

log 140.3log 3 4.5

4.4 a. Solve: 22x – 1 + 3 = 35

22x – 1 = 32

ln 22x – 1 = ln 32

(2x – 1) ln 2 = ln 32

2x – 1 = 5

2x = 6

x = 3

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

4.5 R = 6e12.77x

If R = 25% accident risk

25 = 6e12.77x

25/6 = e12.77x

ln (25/6) = ln e12.77x

ln (25/6) = 12.77x

x = 0.112

A blood alcohol level of 0.112 corresponds to an accident

risk of 25%

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)

F

F

V

V

Center: (0, 0)

The equation must be in the form:

x2 y2 a2 b2

+ = 1

b = distance from the center to a vertex = 4c = distance from the center to a focus pointc = 3

c2 = b2 – a2

9 = 16 – a2

a2 = 7x2 y2 7 16

+ = 1Equation:

7.3 Find the equation of a parabola with vertex at (2, -3) and focus at (2, -5)

F

V

The parabola is vertical with an equation in the form:

(x – h)2 = 4p(y – k)

(h, k ) = (2, -3)

p = -2

Equation: (x – 2)2 = 8(y + 3)