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 Dependent Variable

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# Math 132 TEST #2 Solutions

 Problem 1 2 3 4 Total Possible Score 25 70 40 65 200 Your Score

SHOW ALL WORK. Any solution that is not accompanied by the appropriate work necessary for solving
the problem will receive no credit. If you need more space, you may use the back of the page.

1. (25 pts) Find a polynomial with the following graph. SOLUTION:

The zeros of the polynomial are x = −6, 3.5, and 8, so the polynomial has the form Since the graph bounces off the x-axis at x = 8, we have is even, and since the
graph crosses the x-axis at x = −6 and x = 3.5, we have and are odd.

Since there are 5 turning points, , so let and , so
we have y = k(x + 6)(x − 3.5)(x − 8)4.

To solve for k, we let x = −7 and y = −12 to get Thus, a polynomial would be 2. Let  where A, B, C, and D are real numbers.

(a) (7 pts) What do the y-values approach as x→∞ for j(x) = f(x) · g(x)?

SOLUTION: As so (b) (7 pts) What do the y-values approach as x→−∞ for k(x) = f(x) · h(x)?

SOLUTION: As so (c) (7 pts) What do the y-values approach as x→∞ for ?

SOLUTION: As ,so d) (7 pts) What do the y-values approach as x→−∞ for ?

SOLUTION: As , so Reminder where A, B, C, and D are real numbers.
(e) (15 pts) What are the possible vertical asymptotes of ?

SOLUTION:

The possible vertical asymptotes are the zeros of the denominator, h(x).

Thus, the possible vertical asymptotes have the form (f) (12 pts) What are the possible zeros of ? (Hint: you may need to do some
factoring).

SOLUTION:

The possible zeros are the zeros of the numerator, g(x) = 6x2 + Cx3 − 4x5 + 5x.

Since there is not a constant coefficient, x = 0 is a zero of g(x).

For other possible zeros, we factor an x from g(x) to get g(x) = x(6x + Cx2 − 4x4 + 5).

Therefore, the possible zeros have the form (g) (15 pts) What are the possible holes in the graph of ?

SOLUTION:

The possible holes in the graph are the zeros of both the numerator and denominator.

The possible zeros of h(x) have the form The possible zeros of f(x) have the form So the possible holes have the form 3. Let f(x) = 4x2 + 10x + 9.

(a) (15 pts) Complete the square to write f(x) in the form k(x + h)2 + v.

SOLUTION:

4x2 + 10x + 9 = k(x + h)2 + v = kx2 + 2khx + (h2k + v).

Thus, k = 4, and .

So .
Therefore, (b) (15 pts) Use part (a) to find the x-values such that f(x) = 4.

SOLUTION: (c) (10 pts) What is the smallest y-value of f?
SOLUTION: Since f is a parabola with vertex , the smallest y-value is 4. Let f(x) = 2x3 + 5x2 − 13x + 5 and g(x) = 2x3 − 15x2 + 18x + 35.

(a) (20 pts) Find the zeros of f(x). Hint: you will probably need to use the Rational Zeros
Theorem, long division of polynomials, and the quadratic formula (or completing the square).
You must show all of your work for each of these steps. You may not find the zeros by graphing
the function with your graphing calculator and having the calculator compute the zeros for you.

SOLUTION: The possible rational zeros of f(x) are of the form . , so 2x − 1 is a factor of f(x). By long division, we get So f(x) = (2x − 1)(x2 + 3x − 5).

Using the quadratic formula, the remaining zeros are , so the zeros are Reminder f(x) = 2x3 + 5x2 − 13x + 5 and g(x) = 2x3 − 15x2 + 18x + 35.

(b) (20 pts) Find the zeros of g(x). Hint: you will probably need to use the Rational Zeros
Theorem, long division of polynomials, and the quadratic formula (or completing the square).
You must show all of your work for each of these steps. You may not find the zeros by graphing
the function with your graphing calculator and having the calculator compute the zeros for you.

SOLUTION:

The possible zeros of g(x) have the form .

From these we find g(−1) = 0, g(5) = 0, and .

Thus, the zeros are .

Reminder f(x) = 2x3 + 5x2 − 13x + 5 and g(x) = 2x3 − 15x2 + 18x + 35.

(c) (25 pts) Use parts (a) and (b) to sketch the graph of .
Mark any x-intercepts, y-intercept, horizontal asymptote, vertical asymptotes, and holes. You
must show all of your work on this problem including any work involving asymptotes. SOLUTION:

From parts (a) and (b), we can factor The zeros are x = .5, 1.2,−4.2, and the y-intercept is .

As , so y = 2 is a horizontal asymptote.

As , so x = −1 is a vertical asymptote.

As , so x = 5 is a vertical asymptote.