  # Math Test 2B Solutions

## C. HECKMAN TEST 2A 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

Solution: Its x- intercept is where y = 0, or when , which is
when the numerator is zero . The numerator factors into 2(x − 2)(x − 3), so the
x-intercepts are 2 and 3 .

The y-intercepts are where x = 0, so you just substitute x = 0 into the
formula to get or .

The vertical asymptotes occur where the denominator is zero but the numerator
isn’t. Since the denominator is never zero, there are no vertical asymptotes .

To find the horizontal asymptote, we look at the degrees of the numerator
and denominator. Since both degrees are equal, take the ratio of the coefficients
of x2: or y = 2 .

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

2) Do the following for the equation 4z = 19:

(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 z , rounding your answer to three decimal places.

Solution: You can take logarithms of both sides here, and solve for z, 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] Solution: Since there are multiple logarithms, you need to combine them : Now convert to exponential form : [You got 4 points for getting this far.] If you cross-multiply, you find out that
x + 37 = 10(x + 1) [7 points], so the only possible solution is 3 . It turns out to
be an actual answer, after checking.

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

(b) [10 points] 102x − 10x − 12 = 0

Solution: If you let y = 10x, you find out that [3 points] so y = 4 or y = −3 [7 points]. To find x, you solve the equations
10x = 4 (which has a solution of log 4) and 10x = −3 (which has no solutions).
So log 4 is the only solution.

Grading for partial credit: +7 points for 4 or log(−3). Points were taken off

(4) A bank offers a savings account where the interest is compounded 4.7% monthly.

(a) [5 points] If you deposit \$350 now, how much money will you have in 5 years?

Solution: Use the compound interest formula: Grading: +2 points for the formula, +2 points for substituting, +1 point for

(b) [10 points] When will you have \$800 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 350; the
left-hand side is not 351.3712t. [7 points] [10 points]

Grading for partial credit: −3 points for not dividing by 350 first.

(5) [10 points] Find a polynomial p (x) which has degree 3, has 2 as a root, has −2 as a
root of multiplicity 2, and where p(0) = 16.

Solution: Since you know information about the roots (zeros), it is best to work
with the factored form. Since −2 has multiplicity 2, (x + 2)2 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 4. Hence p(x) = A(x − 2)(x + 2)2
for some constant A.

If you use the condition p(0) = 16, you find out that so A = −2. Thus .

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

(6) [10 points] Find the quotient and remainder when 4x4 +3x3 −4 is divided by x2 −2x.

Solution: Since the divisor is not of the form x − c, you must use long division: The quotient is , and the remainder is .

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 9x4 +5x3 −7x+5. 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 5, and q is a factor of 9. Since
the factors of 5 are 1, 5, −1, and −5, and the factors of 9 are 1, 3, 9, −1, −3,
and −9, we take all possible combinations to get the list below. Grading: +3 points for the factors of 5, +3 points for the factors of 9, 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 −2 and . Find all roots of p(x) exactly, along with their multiplicities.

Solution: First of all, you should divide p(x) by x − (−2), 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 + 38x2 + 34x + 28.

You can’t use the quadratic formula on f(x), since f(x) has degree three.
The instruction to find the multiplicities suggests that −2 or may have multiplicity
greater than one, so you should divide f(x) by x+2 and again, to see
whether one of these divisions produces a remainder of 0. In fact, x+2 divides into
f(x) without a remainder. The quotient of this final division is 14x2 + 10x + 14,
and we can use the quadratic formula to find the remaining roots: .

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

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

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