Test AP Precalculus: unità da 1 a 3

• A. 1
• B. 2
• C. 3
• D. 6

Risposta: B. 2

The average rate of change is the slope of the secant line: (f(4) - f(1)) / (4 - 1). Here f(4) = 16 - 12 = 4 and f(1) = 1 - 3 = -2, so the change in output is 4 - (-2) = 6, and dividing by the change in input, 3, gives 2. The answer 6 forgets to divide, and 3 is the length of the interval.

• A. f(x) approaches negative infinity as x approaches infinity, and f(x) approaches infinity as x approaches negative infinity
• B. f(x) approaches infinity as x approaches infinity, and f(x) approaches negative infinity as x approaches negative infinity
• C. f(x) approaches negative infinity as x approaches either infinity or negative infinity
• D. f(x) approaches infinity as x approaches either infinity or negative infinity

Risposta: A. f(x) approaches negative infinity as x approaches infinity, and f(x) approaches infinity as x approaches negative infinity

For large inputs the leading term, -2x^5, dominates. The degree is odd, so the two ends of the graph go in opposite directions, and the leading coefficient is negative, so the right end goes down. That means f falls toward negative infinity on the right and rises toward infinity on the left. The other terms, 3x^2 and -1, have little influence once |x| is large.

Seleziona tutte le risposte corrette.

• A. The graph has a hole at x = -2
• B. The graph has a vertical asymptote at x = 3
• C. The graph has a vertical asymptote at x = -2
• D. The graph has a horizontal asymptote at y = 1
• E. The graph has an x-intercept at x = -2

Risposta: A. The graph has a hole at x = -2, B. The graph has a vertical asymptote at x = 3, D. The graph has a horizontal asymptote at y = 1

Factor both parts: r(x) = (x - 2)(x + 2) / ((x - 3)(x + 2)). The factor x + 2 cancels, so there is a hole at x = -2, not an asymptote, and not an x-intercept either because r is undefined there. The factor x - 3 is left in the denominator, so there is a vertical asymptote at x = 3. The numerator and denominator have the same degree, so the horizontal asymptote is the ratio of the leading coefficients, 1/1 = 1. The only x-intercept is x = 2.

A cubic polynomial has a positive leading coefficient and zeros at x = -1 with multiplicity 2 and at x = 3 with multiplicity 1. At x = -1 the graph _____ the x-axis, and at x = 3 the graph _____ the x-axis.

Opzioni per lo spazio 1: crosses, touches but does not cross, stays above, stays below

Opzioni per lo spazio 2: crosses, touches but does not cross, stays above, stays below

Risposta: touches but does not cross, crosses

The polynomial is p(x) = a(x + 1)^2 (x - 3) with a > 0. A zero of even multiplicity makes the graph touch the axis and turn back, because the factor (x + 1)^2 does not change sign. A zero of odd multiplicity makes the graph cross. So it touches at x = -1 and crosses at x = 3. The multiplicities add up to 3, which matches a cubic.

• A. (-2, 9)
• B. (4, 8)
• C. (4, 7)
• D. (4, 9)

Risposta: D. (4, 9)

To find the new point, work out which input of f the new graph uses. g(4) = 2f(4 - 3) + 1 = 2f(1) + 1 = 2(4) + 1 = 9, so (4, 9) is on the graph of g. The inside change x - 3 shifts the graph 3 units right, so the x-value goes from 1 to 4. The outside changes stretch the output by 2 and then shift it up 1, so 4 becomes 9. The answer (-2, 9) shifts the wrong way, and the others get the order of the output steps wrong.

A geometric sequence with positive terms has g_3 = 24 and g_5 = 96. Its common ratio is _____ and its initial term g_0 is _____.

Opzioni per lo spazio 1: 2, 3, 4, 12

Opzioni per lo spazio 2: 6, 3, 12, 0.75

Risposta: 2, 3

Going from the 3rd term to the 5th term multiplies by the ratio twice, so r^2 = 96/24 = 4. The terms are positive, so r = 2. The general term is g_n = g_0 times r^n, so g_3 = g_0 times 2^3 = 8 g_0 = 24, which gives g_0 = 3. A common slip is to treat 96/24 = 4 as the ratio itself instead of the ratio squared.

• A. P(t) = 8000(0.05)^t
• B. P(t) = 8000 + 0.05t
• C. P(t) = 8000(5)^t
• D. P(t) = 8000(1.05)^t

Risposta: D. P(t) = 8000(1.05)^t

Growth by 5 percent per year means each year the population is multiplied by 1 + 0.05 = 1.05, which is constant proportional change, so the model is exponential: P(t) = 8000(1.05)^t. The base 0.05 would shrink the population almost to nothing, the base 5 would multiply it fivefold every year, and adding 0.05t is linear growth of 0.05 people per year.

• A. 8
• B. 7
• C. 9
• D. 5

Risposta: A. 8

Rewrite the logarithmic equation in exponential form: log_3(x + 1) = 2 means 3^2 = x + 1. So x + 1 = 9 and x = 8. Check the domain: x + 1 = 9 is positive, so the logarithm is defined. The value 9 is x + 1, not x, and 5 comes from multiplying 3 by 2 instead of raising 3 to the power 2.

Seleziona tutte le risposte corrette.

• A. The domain of f is x > 1
• B. The graph of f has a vertical asymptote at x = 1
• C. f(2) = 0
• D. The inverse function is f^(-1)(x) = e^(x - 2) + 1
• E. f is a decreasing function

Risposta: A. The domain of f is x > 1, B. The graph of f has a vertical asymptote at x = 1, D. The inverse function is f^(-1)(x) = e^(x - 2) + 1

The logarithm is defined only when x - 1 > 0, so the domain is x > 1, and the graph approaches the vertical line x = 1 without touching it. To find the inverse, solve y = ln(x - 1) + 2 for x: subtract 2 and exponentiate to get x - 1 = e^(y - 2), so x = e^(y - 2) + 1. Then swap the roles of x and y, which gives f^(-1)(x) = e^(x - 2) + 1. For the other statements, f(2) = ln(1) + 2 = 2, not 0, and a natural logarithm with a positive coefficient is increasing, so f is not decreasing.

• A. The model is appropriate, because the residuals are small
• B. The model is appropriate, because the residuals fall both above and below the axis
• C. The exponential model is not appropriate, because residuals from an appropriate model show no pattern
• D. The data must be periodic, so a sinusoidal model is required

Risposta: C. The exponential model is not appropriate, because residuals from an appropriate model show no pattern

A residual is the actual value minus the value the model predicts. If a model fits well, its residuals scatter randomly around zero with no visible pattern. A curve, a trend or a fan shape means the model is missing some structure in the data, so it is not appropriate. Residuals landing on both sides of the axis happens even for bad models, and small residuals alone do not rule out a systematic pattern. A bad fit does not by itself mean the data are periodic.

• A. pi/3 and 5pi/3
• B. pi/3 and 2pi/3
• C. pi/6 and 11pi/6
• D. pi/3 only

Risposta: A. pi/3 and 5pi/3

Divide by 2 to get cos(theta) = 1/2. On the unit circle the x-coordinate is 1/2 at two angles in one turn: pi/3 in the first quadrant and 5pi/3 in the fourth. The answer 2pi/3 has cosine -1/2, and pi/6 and 11pi/6 have cosine sqrt(3)/2. Always look for the second solution, because the cosine takes most values twice per period.

The function f(theta) = 3sin(2theta) + 1 has amplitude _____ and period _____.

Opzioni per lo spazio 1: 1, 2, 3, 6

Opzioni per lo spazio 2: pi/2, pi, 2pi, 4pi

Risposta: 3, pi

In a form like a sin(b(theta - c)) + d, the amplitude is the absolute value of a, here 3, and the period is 2pi divided by the absolute value of b, here 2pi/2 = pi. The +1 moves the midline up to y = 1 but does not change the amplitude, so the function oscillates between -2 and 4. A larger b squeezes the graph horizontally, so doubling the input halves the period from 2pi to pi.

• A. h(t) = 18cos(pi t / 6) + 22
• B. h(t) = -18cos(pi t / 6) + 22
• C. h(t) = -18cos(pi t / 12) + 22
• D. h(t) = -22cos(pi t / 6) + 18

Risposta: B. h(t) = -18cos(pi t / 6) + 22

The midline is the average of the maximum and minimum, (40 + 4)/2 = 22, and the amplitude is half the range, (40 - 4)/2 = 18. The period is 12, so b = 2pi/12 = pi/6. A plain cosine starts at its maximum, but this rider starts at the minimum, so the cosine is reflected with a negative sign: h(0) = -18 + 22 = 4. The version with pi/12 would have a period of 24 minutes, and the last option swaps the amplitude and the midline.

Seleziona tutte le risposte corrette.

• A. (3, 9pi/4)
• B. (-3, 5pi/4)
• C. (3, 3pi/4)
• D. (-3, pi/4)
• E. (3, -7pi/4)

Risposta: A. (3, 9pi/4), B. (-3, 5pi/4), E. (3, -7pi/4)

A point has many polar names. Adding or subtracting 2pi to the angle gives the same point, so (3, pi/4 + 2pi) = (3, 9pi/4) and (3, pi/4 - 2pi) = (3, -7pi/4) both work. A negative radius puts the point on the opposite side of the origin, so the same point can be named with the opposite direction, pi/4 + pi = 5pi/4, and radius -3. The point (3, 3pi/4) is a different location, in the second quadrant, and (-3, pi/4) is the reflection of the original point through the origin.

• A. It increases from 1 to 4
• B. It decreases from 4 to 1
• C. It increases from -4 to -1
• D. It stays constant

Risposta: B. It decreases from 4 to 1

The distance from the origin is the absolute value of r, because a negative r only flips the direction of the point and not its distance. As r goes from -4 up to -1, |r| goes from 4 down to 1, so the point moves closer to the origin even though r itself is increasing. A distance cannot be negative, which rules out the third option. The same idea works for a positive r that decreases: the distance shrinks.