Algebra: Functions
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1. The function y=f(x) is shown above. Which of the diagrams below represents the function y=2·f(x+90)+2?

Note: The answer choices do not all use the same scale.

The f(x+90) means shift left by 90 (which makes x=0, y=1 a point on the new graph). The 2*f(x+90) means double everything so it is twice as tall in both directions (x=0, y=2 is now a point on the graph). The 2*f(x+90) + 2 means raise the whole thing by 2 (x=0, y=4 is a point on the final graph). The answer is (A).

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(A)
(B)
(C)
(D)
(E)


2. Which of the diagrams below is the graph of the function y = f(x) = x3+5x2+6x?

Factor the function to get f(x) = x(x2 + 5x + 6) = x(x+2)(x+3). So f(x) is zero when x is 0, —2, or —3. That means the answer is (C).

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(A)

(B)

(C)

(D)

(E)


3. In the diagram above, the functions y=f(x) and y=g(x) intersect at points (0,0) and (a,b). The line x=a is also shown. If c is a number between 0 and a such that 0 ≤ c ≤ a, then which of the following must be true?

I. f(c) – g(c) ≥ 0
II. f(a/2) > g(c)
III. The area of triangle ABC < a·f(a)

(A) I, only
(B) II, only
(C) I and II, only
(D) II and III, only
(E) I and III, only

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The function f is always above the the function g (except at the end points where they are equal) so I is true. If you bring your finger to a/2 and then go up to f(a/2) you get pretty high but g(c) could be even higher depending on what c is (it can be anything between 0 and a) so II is NOT true. Since a is the base of the triangle and f(a) is the height of the triangle, and the area is only half the base times the height, III is certainly true. So the answer is (E).

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4. In the diagram above, line p intersects the curve y = √x at point B and line BC is perpendicular to line AC. The length of line segment AB is √6. What is the area of triangle ABC?

(A) √2
(B) 2 · √2
(C) √3
(D) 2 · √3
(E) cannot be determined

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It looks impossible, but it isn't. Since it's an intersection problem you know point B is (x, y) or (x, √x). The pythagorean theorem (SATan's favorite) gives you (x)2 + (√x)2 = (√6)2 or x2+x—6=0 or (x+3)(x—2) = 0. Since x can't be negative we get x = 2 so point B is (2, √2). So the base of the triangle is 2 and the height is √2. This means the area is √2 and the answer is (A).

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5. If f(x) = x2 + x + 2, then which of the equations below has the same solution set as f(x+2) = 4?

(A) x2 + 5x + 2 = 0
(B) x2 + 5x + 4 = 0
(C) x2 + 5x + 6 = 0
(D) x2 + 3x + 2 = 0
(E) x2 + 3x + 6 = 0

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Your equation is f(x+2) = (x+2)2 + (x+2) + 2 = 4. If you do your foiling and combine terms you get x2 + 5x + 8 = 4 which is the same as choice (B). Don't let that "same solution set" stuff bother you; translate it to: "Which equation is the same as f(x+2) = 4."

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6. If f(x) = 4x2 + 6, for what value of x does (1/2) · f(x) = f(2x)?

(A) 0
(B) 3/14
(C) √(3/14)
(D) –√(3/14)
(E) no solution

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Just plug in. Half of f(x) is 2x2 + 3 and f(2x) is 16x2 + 6. Setting these equal gives you 14x2 = —3 which has no solution because x-squared is always positive. The answer is (E).

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7. A line with slope m intersects the y-axis at y = 2 and intersects the curve y = x2–2 at point (3,b). What is the value of m?

(A) 5/3
(B) 7/3
(C) 2/9
(D) 5/9
(E) 7/9

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"Intersects the y-axis" means x = 0 so one of our points is (0,2). The other point is the usual (x, f(x)) that you see in all intersection problems. In this case (x, f(x)) or (x, y) is (3, 32-2) or (3,7). Now all we have to do is rise (5) over run (3) and the answer is (A).

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8. Two curves, y = px2 and y = qx2 intersect rectangle ABCD at its four corners. Which of the following expressions gives the area of the rectangle?

(A) 4 · (p + q)
(B) 8 · (p + q)
(C) 16 · (p + q)
(D) 8 · (p – q)
(E) 16 · (p – q)

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The intersection points are (x, y) = (x, f(x)) as always. Point C is (2, 4p) and Point B is (2, 4q). Note that q is negative. Side CB = 4p - 4q and side AB = 4. So the area is 16·(p-q) or (E).

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9. The curve y = 2x2 intersects equilateral triangle ABC at the origin at vertex A and at vertex B. What is the area of the triangle?

(A) √3
(B) 3 · √3
(C) 3 · √3/2
(D) 3 · √3/4
(E) cannot be determined

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It looks impossible but it isn't. The key to any intersection problem is to realize that the point of intersection, point B, is always (x, f(x)) which is (x, 2x2) in this case. Since the triangle is equilateral, each half is 30-60-90 which means that (√3)x = 2x2. Solving this gives x = √3/2 which means point B is (√3/2, 3/2). Now you can get the area of the triangle. The area of the triangle is one-half base times height or 3·√3/4. So the answer is (D).

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10. A fancy car containing a number of (fictional) ETS executives and their mistresses is moving along a country road at constant speed. The driver is drunk and hits a tree causing the car to stop suddenly (there are several minor injuries and one arrest). Which distance (x) vs. time (t) graph best represents the motion of the car?

You have to be careful with these. When it stops suddenly that means the speed goes to zero, NOT the distance (x). So choices, A, C, and D are all wrong. A constant speed means (x) increases steadily so the graph is a diagonal line. When the car hits the tree, (x) stops increasing and "flatlines." Choice E would be right if the car had bounced backwards after hitting the tree. It just "stopped suddenly" so the answer is (B).

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(A)
(B)
(C)
(D)
(E)