1. Graph the following functions. Show all your work!

(a) f(x) = x(x − 1)(x + 2)

Solution:

End behavior:

As x → ∞ f(x) → (+)(+)(+) = +∞

As x → −∞ f(x) → (−)(−)(−) = −∞

y-intercept: 0

x-intercepts: 0, 1, −2 (all multiplicity 1)

Hence the graph is

(b) f(x) = (x + 3)^{2}(2 − x)(x + 1)^{3}

Solution:

End behavior:

As x → ∞ f(x) → (+)^{2}(−)(+)^{3} = −∞

As x → −∞ f(x) → (−)^{2}(+)(−)^{3} = −∞

y-intercept: 18

x-intercepts: −3 (multiplicity 2), 2 (multiplicity 1), −1 (multiplicity

3)

Hence the graph is

2. Find the quotient and remainder for

Solution:

Hence the quotient is x^{3} + 4x^{2} + 4x + 2 with a remainder
of 5x − 1.

3. Find f(−5) if f(x) = x^{5} + 8x^{4} + 12x^{3} − 14x^{2} − 22

Solution:

Use the remainder theorem and synthetic division:

Hence f(−5) = 3

4. Find all zeros of the function. Show all your work!

(a) f(x) = x^{3} − 4x^{2} − 3x + 18

Solution:

List the possible rational zeros: ±1,±2,±3,±6,±9,±18

Clearly 1 and −1 are not zeros, so we try 2:

Not a zero, so we try −2:

This is a zero, so we can factor the polynomial as
(x+2)(x−6x+9)

We can now factor the quadratic term to get: (x + 2)(x − 3)^{2}

Hence the zeros are −2 and 3 (multiplicity 2).

(b) f(x) = x^{4} − 2x^{3} − 7x^{2} + 20x − 12

Solution:

First list the possible rational zeros: ±1,±2,±3,±4,±6,±12

First check 1:

Hence we can factor the polynomial as (x − 1)(x^{3} − x^{2} − 8x
+ 12)

We have to find another zero, the possible rational zeros
remain the

same. It is easy to see that neither 1 or −1 is a zero, so we try 2:

Hence 2 is a zero and we can factor the polynomial as (x −
1)(x −

2)(x^{2} + x − 6)

Factoring the quadratic term we get (x − 1)(x − 2)(x − 2)(x + 3)

Hence the zeros are 1, 2 (multiplicity 2), −3

5. Graph the function . Explain all your
steps!

Solution:

First factor the polynomials:

Horizontal asymptote: y = 0 (degree smaller in numerator)

Vertical asymptotes: 0 (multiplicity 1), 2 (multiplicity 2)

y-intercept: none

x-intercept: −1 (multiplicity 2)

Steps to graph it: (from right to left)

First check the sign to the right off the vertical asymptote at x = 2.

It is positive, so the graph will be above the x- axis.

Thus it will go up on the right of the vertical asymptote at x = 2.

The vertical asymptote at x = 2 has multiplicity 2, so there will not be

a change in sign. Thus it is also going up to the left of the asymptote.

Now there is the vertical asymptote at x = 0. Since there is no x-

intercept between the two asymptotes, it will have to go up to the right

of this asymptote.

This asymptote has multiplicity 1, so there is a sign change. Thus to

the left of the asymptote at x = 0 it will go down.

Now there is the x-intercept at −1. It has multiplicity 2, so there will

not be a sign change. Hence to the left of this intercept it will also be

negative.

Thus the graph is:

6. Find the slant asymptote of
and graph the function.

Solution:

First find the slant asymptote by synthetic division:

Hence the function is

Thus the slant asymptote is y = 3x + 2

Factor the numerator: Vertical asymptote: x
= −1

y-intercept: −2

x-intercepts: , −2

Then the graph is