After a brief hiatus, the Algebra 2 Problems of the Day are back. Hopefully, daily.
More Algebra 2 problems.
January 2019, Part I
All Questions in Part I are worth 2 credits. No work need be shown. No partial credit.
13. The function f(x) = a cos bx + c is plotted on the graph shown below.
What are the values of a, b, and c?
(1) a = 2, b = 6, c = 3
(2) a = 2, b = 3, c = 1
(3) a = 4, b = 6, c = 5
(4) a = 4, b = π / 3, c = 3
Answer: (1) a = 2, b = 6, c = 3
Because c is not in parentheses, it is a vertical shift, not horizontal, and it will be equal to the midline of the function, which is 3 (halfway between 1 and 5). This eliminates choices (2) and (3).
Next, a is the amplitude it is equal to the distance from the midline to the maximum point (or one-half the distance from the maximum to the minimum). So a = 2, not 4. Eliminate choice (4).
As for b, it tells us that there will be 6 cycles in a 2π interval, and 2π / 6 = π / 3, which is the period for one cycle.
14. Which equation represents the equation of the parabola with focus
(-3,3) and directrix y = 7?
(1) y = 1/8 (x + 3)2 - 5
(2) y = 1/8 (x - 3)2 + 5
(3) y = -1/8 (x + 3)2 + 5
(4) y = -1/8 (x - 3)2 + 5
Answer: (3) y = -1/8 (x + 3)2 + 5
The standard form of a parabola is y = a(x − h)2 + k, with the vertex at (h, k) and the focus at (h, k + 1 / (4a)).
Since the vertex is equidistant between the focus and the directrix, then vertex must be at (-3, 5) because 5 is halfway between 3 and 7. (3 + 7) / 2 = 5.
Flipping the sign, we know that (x + 3) must be in the equation, so eliminate (2) and (4).
Also, the equation ends with + 5, so we can eliminate choice (1) as well.
At this point, we know the answer is (3). We have one more piece of information: because the directrix is above the vertex, which is above the focus, we know that the parabola opens downward, so a must be negative:
1 / ((4)(-2)) = a
1 / (-8) = a
15. What is the solution set of the equation
(1) { -1/3, 1/2 }
(2) { -1/3 }
(3) { 1/2}
(4) { 1/3, -2 }
Answer: (3) { 1/2}
Strategy: find a common denominator by multiplying each fraction by either (x / x) or (3x + 1)/(3x + 1).
Then eliminate all the denominators by multiplying both sides of the equation by (x)(3x + 1).
However, we will have to check for extraneous solutions. Because of the denominators in the original equations, we know that x =/= 0 and 3x + 1 =/= 0.
2x / ((3x + 1)(x)) = 1 / x - 6x(x) / ((3x + 1)(x))
2x / ((3x + 1)(x)) = (3x + 1) / ((3x + 1)(x)) - 6x(x) / ((3x + 1)(x))
2x = 3x + 1 - 6x2
6x2 - x - 1 = 0
(3x + 1)(2x - 1) = 0
3x + 1 = 0 or 2x - 1 = 0
3x = -1 or 2x = 1
x = -1/3 or x = 1/2
However, 3x + 1 =/= 0, and 3(-1/3) + 1 = 0, so we eliminate that answer. That leaves only 1/2, which is Choice (3).
Comments and questions welcome.
More Algebra 2 problems.
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