Math Test 2B Solutions

C. HECKMAN TEST 2B
SOLUTIONS 170

(1) [10 points] For the rational function below, find its x-intercept(s), y-intercept(s), vertical
asymptotes, horizontal asymptotes, and determine whether it is even, odd, or
neither. (If the function does not have any vertical asymptotes (for example), make
sure you answer “none”; do not leave your answer blank.)

Solution: Its x-intercept is where y = 0, or when , which is
when the numerator is zero . The numerator factors into 3(x + 4)(x − 1), so the
x-intercepts are −4 and 1 .

The y-intercepts are where x = 0. If you substitute x = 0 into the formula,
you have to divide by zero , so this function has no y-intercept .

The vertical asymptotes occur where the denominator is zero but the numerator
isn’t. Since the denominator is zero when x³(x + 5) = 0, the vertical
asymptotes are x = 0 and x = −5 .

To find the horizontal asymptote, we look at the degrees of the numerator
and denominator. Since the degree of the denominator is greater, there is a
horizontal asymptote of y = 0 .

Since , which is neither the original function nor its
negative , this function is neither even or odd.

Grading: +2 points for each answer. Partial credit: +1 point for each answer.

Do the following for the equation 2^y = 5:

(a) [5 points] Convert it into logarithmic form .

Solution: .

Grading for partial credit: +2 points total for taking logarithms of both sides.

(b) [10 points] Solve for y , rounding your answer to three decimal places .

Solution: You can take logarithms of both sides here, and solve for y, or use the
change of basis formula:

Note that will also give you the correct answer.

(3) Solve the following equations for x . You must give exact answers!

(a) [10 points] ln(x + 1) + ln(2x − 1) = 3

Solution: Since there are multiple logarithms , you need to combine them :

Now convert to exponential form :

so you need to solve the quadratic equation

[You got 4 points for getting this far.] The solutions to this quadratic equation
are [7 points], but only the solution with the + sign makes
the original equation true. So the only solution is . (This is not
3, since e³ is not exactly 20 but only approximately 20.)

Grading for partial credit: +3 points for miscellaneous work.

(b) [10 points]

Solution: If you let y = 10^x, you find out that

[3 points] so y = 1 or y = 4 [7 points]. To find x, you solve the equations 10^x = 1
(which has a solution of 0) and 10^x = 4 (which has a solution of log 4). Both
solutions work in the original equation, so the solutions are 0 and log 4 .

Grading for partial credit: +7 points for 4 and 1. Points were taken off for
an approximate answer.

(4) A bank offers a savings account where the interest is compounded 5.1% quarterly.

(a) [5 points] If you deposit $300 now, how much money will you have in 4 years?

Solution: Use the compound interest formula:

Grading: +2 points for the formula, +2 points for substituting, +1 point for
the final answer. Grading for partial credit: −2 points for .

(b) [10 points] When will you have $400 in your account?

Solution: You need to solve the equation

[3 points]

To solve this equation, you have to start by dividing both sides by 300; the
left-hand side is not 303.825^4t.

[7 points]

[10 points]

Grading for partial credit: −3 points for not dividing by 300 first.

(5) [10 points] Find a polynomial p (x) which has degree 4, has 2 as a root, has −1 as a
root of multiplicity 3, and where p(1) = −24.

Solution: Since you know information about the roots (zeros), it is best to work
with the factored form. Since −1 has multiplicity 3, (x + 1)³ is a factor of p(x),
and (x−2) is also a factor of p(x). Note that if the multiplicity of 2 were greater
than 1, then p(x) would have degree at least 5. Hence p(x) = A(x − 2)(x + 1)³
for some constant A.

If you use the condition p(1) = −24, you find out that

so A = 3. Thus .

Grading for partial credit: +3 points (total) for involving (x − 2)(x + 1);
+7 points (total) for (x − 2)(x + 1)³.

(6) [10 points] Find the quotient and remainder when 3x⁴−7x²+8x+5 is divided by x²+2.

Solution: Since the divisor is not of the form x − c, you must use long division:

The quotient is 3x² − 13 , and the remainder is 8x + 31 .

Grading was done on a 0–3–5–7–10 point basis.

(7) [10 points] Using the Rational Root Theorem, list the possible rational roots of the
polynomial 3x⁴ −8x² +5x−4. You do not need to determine which are actual roots.

Solution: The Rational Root Theorem states that all rational roots of this polynomial
are of the form , where p is a factor of −4, and q is a factor of 3. Since
the factors of −4 are 1, 2, 4, −1, −2, and −4, and the factors of 3 are 1, 3, −1,
and −3, we take all possible combinations to get the list below.

Grading: +3 points for the factors of −4, +3 points for the factors of 3, and
+4 points for combining them. Grading for common mistakes: +8 points (total)
if the reciprocals were given.

(8) [10 points] The rational roots of the polynomial

are −1 and. Find all roots of p(x) exactly, along with their multiplicities.

Solution: First of all, you should divide p(x) by x − (−1), and then divide that
quotient by , to try to find an equation that the other roots must satisfy.
Doing this division (by long division or synthetic division) yields f(x) =
14x³ + 24x² + 4x − 6.

You can’t use the quadratic formula on f (x), since f(x) has degree three.
The instruction to find the multiplicities suggests that −1 or may have multiplicity
greater than one, so you should divide f(x) by x+1 and again, to see
whether one of these divisions produces a remainder of 0. In fact, x + 1 divides
into f(x) without a remainder. The quotient of this final division is 14x²+10x−6,
and we can use the quadratic formula to find the remaining roots: .

The roots of p(x) are thus −1 (with multiplicity 2, since you divided by x+1
twice), , and .

Grading: +5 points for finding f(x), +2 points for dividing again by x + 1,
+3 points for using the quadratic formula. Grading for partial credit: +5 points
(total) for −1 and .