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Math 117 Midterm Review 2

First part, free responses.

I. Solve the following problems.

(f) Find the principle square root of −16.

II. Suppose f(x) = (x + 9)(x − 2)2.

(a) Find all x- and y-intercpets.
Setting f(x) = 0, and solving for x, we obtain x- intercepts at points (−9, 0),
(2, 0).

Also in order to obtain y-intercpet, we need to solve f(0). f(0) = (0 +
9)(0 − 2)2 = 36. Therefore y-intercpet is at the point (0, 9).

(b) Determine whether graph touches or crosses x-axis at each x-intercept.
The real zero −9 has a multiplicity of 1, an odd interger, meaning graph
there crosses x-axis.

The real zero 2 has a multiplicity of 2, an even integer, meaning graph there
touches x-axis.

(c) For large value of |x|, the graph will resemble the power function y = x3.

III. Let f(x) = (a − 1)x2 + 4x + 1.

(a) When a−1 < 0, the parabola opens down, so (−∞, 1) should be the answer.

(b) In order for the quadratic function to have two distinct real zeros, the
discriminant need to be positive, i.e., Δ = 42−4·(a−1)·1 = 16−4(a−1) =
20 − 4a > 0. Solving this inequality, we obtain the solution in interval
notation, (−∞, 5).

(c) When a = 5, f(x) = 4x2 + 4x + 1 = (2x + 1)2 ≥ 0, so the domain for this
quadratic function is (−∞,∞), and the range is [0,∞).

IV. Let f(x) = x2 + 4x − 5.

(a) Completing the squares, we have f(x) = (x2+4x+4−4)−5 = (x+2)2−9.
Hence the vertex should be (−2,−9).

(b) Factoring the polynomial function, f(x) = (x−1)(x+5). So setting f(x) =
0, we obtain the real zeros of f are 1 and −5.

(c) Following part (b), we divide the real number line into 3 sections : (−∞,−5),
(−5, 1) and (1,∞). Setting up a table might clear things up:

So, in order for f(x) > 0, the solution intervals are (−∞,−5) ∪ (1,∞).

V. (a) R(p) = p · x = p(−5p + 100) = −5p2 + 100p.

(b) By completing the squares, R(p) = −5(p2 − 20p + 102 − 102) = −5(p −
10)2 + 500. So the vertex is at (10, 500), meaning when price p equals 10,
the quadratic function R(p), revenue, reaches its local maximum of 500.

VI. Suppose .

(a) The domain excludes x = −3, 4, so domain of f(x) is (−∞,−3) ∪ (−3, 4) ∪
(4,∞).
(b) Notice that f(x) is not in lowest terms , so we reduce it into .
Then setting the denominator to 0, we obtain the vertical asymptote x = 4.

(c) We can use long division to obtian the horizontal asymptote, which is y = 2.

VII. Solve inequalities.

(a) x2 − 5x + 4 ≥ 0

x2 −5x+4 = (x−1)(x−4) = 0 → x = 1, 4. So we separate the x-axis into
three parts. Setting up a table:

Hence the solution for f(x) > 0 is (−∞, 1) ∪  (4,∞). Also notice that the
inequality is not strict , we need add solutions for f(x) = 0 into the result,
so we obtain the result as (−∞, 1]∪ [4,∞)



First of all we subtract both sides by 1 and simplify the ineequality, then
we obtain . So the result is (−∞, 1).

Second part, multiple choices .

1. C.

2. C. By looking at the discriminants.

3. C. Using synthetic division twice, we can obtain the answer.

4. A.

5. E.

6. B.

7. D.

8. B.

9. A.

10. C.

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