AP Precalculus — Pattern-Matched Mock Exam

The average rate of change between two points is the change in the output (the y-values) divided by the change in the input (the x-values). Think of it as the slope of the line connecting those two points on the graph. Choice A is just the total change in temperature, not the rate. Choice C is the average temperature, not the rate of change. Choice D has the numerator reversed, which would give you the negative of the correct rate.

This is a function composition problem. You work from the inside out. First, find f(2) = 3(2) - 5 = 6 - 5 = 1. Now, take that result and plug it into g(x). So, we need to find g(1) = (1)^2 + 1 = 1 + 1 = 2. A common mistake is to calculate f(g(2)), which would be f(2^2+1) = f(5) = 3(5)-5 = 10 (Choice C), or to just multiply the functions.

To find the inverse of a function, you swap x and y and solve for the new y. Starting with y = 2^x, we swap to get x = 2^y. To solve for y, we use the definition of a logarithm: if b^y = x, then y = log_b(x). In our case, this becomes y = log_2(x). The other choices represent common confusions: Choice A is a reflection, Choice D is the reciprocal, and Choice B confuses exponential functions with power functions.

Coterminal angles are angles that share the same terminal side. To find them, you add or subtract full rotations, which is 2π radians. So, we're looking for π/4 + 2πn for some integer n. Let's check the options. π/4 + 2π = π/4 + 8π/4 = 9π/4. So, Choice D is correct. The other options are different angles in different quadrants (or a reflection across the x-axis for -π/4).

Let's plug in the points. From (1, 5), we get 5 = k * 1^p, which simplifies to k=5. Now we know f(x) = 5 * x^p. Using the second point (2, 40), we get 40 = 5 * 2^p. Divide by 5: 8 = 2^p. Since 2^3 = 8, the value of p must be 3. This tests your ability to work with function definitions and solve for parameters.

To find vertical asymptotes and holes, you must first factor the numerator and denominator. h(x) = ((x-2)(x+2)) / ((x+2)(x+3)). Notice the (x+2) term appears in both. This means it cancels out, and we have a hole in the graph at x=-2. The (x+3) term is left in the denominator and does not cancel. This creates a vertical asymptote at x=-3. A common mistake is to say there are asymptotes at both x=-2 and x=-3, but you must check for canceling factors first!

For end behavior of rational functions, we compare the degrees of the numerator and the denominator. The numerator's degree is 4, and the denominator's degree is 2. Since the degree of the numerator is greater than the degree of the denominator (n > m), there is no horizontal asymptote. The end behavior is determined by the leading terms: (3x^4) / (5x^2) = (3/5)x^2. The function behaves like y = (3/5)x^2 for large |x|. Since x^2 is always positive, the function goes to positive infinity on both ends. Students often mistakenly choose y=3/5 (Choice A) by just looking at the leading coefficients, but that rule only applies when the degrees are equal.

First, let's solve the equation. Using the product rule for logs, we get log_3((x+5)(x-1)) = 3. Now, convert to exponential form: (x+5)(x-1) = 3^3. This gives x^2 + 4x - 5 = 27, or x^2 + 4x - 32 = 0. Factoring this gives (x+8)(x-4) = 0, so the proposed solutions x=-8 and x=4 are correct algebraically. However, we must check for extraneous solutions. The domain of a logarithm requires the argument to be positive. If we plug x=-8 into the original equation, we get log_3(-3) and log_3(-9), which are undefined. If we plug in x=4, we get log_3(9) and log_3(3), which are both valid. So, only x=4 is a true solution.

Let t be the number of hours past 1:00 PM. The population P(t) can be modeled by P(t) = P_0 * b^t. We know P(0) = 1000. At 3:00 PM, t=2, and P(2) = 4000. So, 4000 = 1000 * b^2. Dividing by 1000 gives 4 = b^2, so the growth factor b is 2. The model is P(t) = 1000 * 2^t. We want to find the population at noon, which is one hour before 1:00 PM, so t=-1. P(-1) = 1000 * 2^(-1) = 1000 * (1/2) = 500. A common mistake is to think linearly; you might see that the population increased by 3000 in 2 hours, so it must have increased by 1500 in the previous hour, leading to 1000-1500 = -500, which is nonsensical. Exponential growth means it multiplies by the same factor each hour.

The key to simplifying trig expressions is often to convert everything to sine and cosine. sec(θ) = 1/cos(θ) and tan(θ) = sin(θ)/cos(θ). Substituting these in gives: ( (1/cos(θ)) - cos(θ) ) / (sin(θ)/cos(θ)). Let's work on the numerator first by finding a common denominator: (1/cos(θ)) - (cos^2(θ)/cos(θ)) = (1 - cos^2(θ)) / cos(θ). We know the Pythagorean identity sin^2(θ) + cos^2(θ) = 1, which rearranges to sin^2(θ) = 1 - cos^2(θ). So the numerator is sin^2(θ) / cos(θ). Now our full expression is ( sin^2(θ) / cos(θ) ) / ( sin(θ) / cos(θ) ). To divide by a fraction, we multiply by the reciprocal: (sin^2(θ) / cos(θ)) * (cos(θ) / sin(θ)). The cos(θ) terms cancel, and one sin(θ) cancels, leaving just sin(θ).

Let's break this down. The amplitude a is half the distance between the max and min: a = (10 - (-2)) / 2 = 12 / 2 = 6. The midline d is the average of the max and min: d = (10 + (-2)) / 2 = 8 / 2 = 4. The period is related to b by the formula Period = 2π / |b|. We are given the period is π. So, π = 2π / |b|. Solving for b gives |b| = 2. Matching these values (a=6, b=2, d=4) with the choices, we find that Choice A is the correct one. It's easy to mix up the formulas, for example, by thinking the period is b (Choice D) or calculating the amplitude or midline incorrectly.

Let's analyze the end behavior. The graph starts low (in Quadrant III, x -> -∞, y -> -∞) and ends high (in Quadrant I, x -> ∞, y -> ∞). When the ends go in opposite directions, the degree of the polynomial must be odd. When it rises from left to right, the leading coefficient must be positive. This matches Choice B. We can also see from the roots: there's a single root at x=-2 and a double root (a 'touch and turn') at x=1. This suggests a minimum degree of 1+2=3, which is an odd number.

A semi-log plot is specifically designed to test for exponential relationships. If a set of data (x, y) becomes linear when you plot (x, log(y)), it means the original relationship was exponential. We can see this by converting the given linear equation back: log(y) = 0.5x + 2. If we assume a base-10 log, then y = 10^(0.5x + 2). This can be rewritten as y = 10^2 * 10^(0.5x) = 100 * (10^0.5)^x. This is clearly in the form y = a * b^x, which is an exponential function.

Let's work from the inside out. Let θ = arcsin(-1/2). This means sin(θ) = -1/2 and θ must be in the range of arcsin, which is [-π/2, π/2]. The angle in that range where sin(θ) = -1/2 is θ = -π/6. Now we need to find cos(θ), which is cos(-π/6). Since cosine is an even function, cos(-π/6) = cos(π/6) = sqrt(3)/2. Alternatively, you can draw a right triangle in Quadrant IV with opposite side -1 and hypotenuse 2. The adjacent side would be sqrt(2^2 - (-1)^2) = sqrt(3). Then cos(θ) is adjacent/hypotenuse, which is sqrt(3)/2.

The inverse function f⁻¹(y) gives you the input x that produced the output y. So, f⁻¹(8) is asking, 'For what x value is f(x) equal to 8?'. Looking at the table, we scan the f(x) row for the value 8. We find it in the third column. The corresponding x value in that column is 6. Therefore, f⁻¹(8) = 6. A common mistake is to find f(8), which would be 10 (Choice A).

This is a transformation problem with two steps. The (x+2) inside the function causes a horizontal shift. Remember, it's the opposite of what you might think: +2 shifts the graph 2 units to the left. The negative sign outside the function, -f(...), causes a reflection across the x-axis, flipping the V-shape to open downwards. Applying these two transformations to the original V-shape (vertex at (0,0)) results in a new V-shape that opens downwards and has its vertex at (-2, 0).

The first sequence, a_n = 3 + (n-1) * 4, is in the form of an arithmetic sequence, a_n = a_1 + (n-1)d. Here, you start at 3 and add a common difference of 4 each time. This corresponds to a constant rate of change (like a linear function). The second sequence, b_n = 3 * 4^(n-1), is in the form of a geometric sequence, b_n = b_1 * r^(n-1). Here, you start at 3 and multiply by a common ratio of 4 each time. This corresponds to a constant percent rate of change (like an exponential function). Therefore, Choice C is the correct description.

Since we know x=2 is a zero, (x-2) must be a factor of the polynomial. We can use synthetic division or polynomial long division to divide p(x) by (x-2). Using synthetic division with 2: 1 -2 1 -2 becomes 1 0 1 0. The resulting quotient is 1x^2 + 0x + 1, or x^2 + 1. To find the other zeros, we set this quotient to zero: x^2 + 1 = 0. This gives x^2 = -1, so the other two zeros are x = i and x = -i. This shows that complex zeros of polynomials with real coefficients always come in conjugate pairs.

Let u = 2x. The equation becomes sin(u) = 0.5. If x is in [0, 2π], then u = 2x must be in the interval [0, 4π]. We are looking for how many times the sine function equals 0.5 over two full rotations of the unit circle. In the first rotation ([0, 2π]), sin(u) = 0.5 at u = π/6 and u = 5π/6. In the second rotation ([2π, 4π]), the solutions are u = π/6 + 2π = 13π/6 and u = 5π/6 + 2π = 17π/6. So we have four solutions for u. Since each u gives a unique x = u/2 that is within the original interval [0, 2π], there are 4 solutions for x. The period of sin(2x) is π, so it completes two full cycles in [0, 2π], crossing the line y=0.5 twice in each cycle.

The conversion formulas are x = r * cos(θ) and y = r * sin(θ). Plugging in our values: x = -3 * cos(π/2) = -3 * 0 = 0. And y = -3 * sin(π/2) = -3 * 1 = -3. So the Cartesian coordinates are (0, -3). You can also think about this visually: θ = π/2 is the positive y-axis. A radius of -3 means you go 3 units in the opposite direction, which is down the negative y-axis to the point (0, -3).

f is a linear function since it has a constant rate of change. Its equation is f(x) = 2x + 10. At x=10, f(10) = 2(10) + 10 = 30. g is an exponential function since it has a constant percent rate of change. Its equation is g(x) = 10 * (1.20)^x. At x=10, g(10) = 10 * (1.20)^10. We don't need to calculate the exact value, just recognize that (1.2)^10 is a number significantly larger than 1 (it's about 6.19). So g(10) will be approximately 61.9. A key concept is that exponential growth, even with a small-seeming percentage, will always eventually overtake linear growth.

This problem uses the rules of logarithms. First, the logarithm of a quotient is the difference of the logarithms: log(x^3) - log(sqrt(y)). Next, the logarithm of a power is the exponent times the logarithm. Remember that sqrt(y) is the same as y^(1/2). So, log(x^3) becomes 3log(x) and log(y^(1/2)) becomes (1/2)log(y) or 0.5log(y). Putting it all together, we get 3log(x) - 0.5log(y). Choice D is a very common mistake, confusing the log of a quotient with the quotient of the logs.

The rate of change of a polar function r = f(θ) refers to how the radius r changes as the angle θ changes. If the rate of change dr/dθ is positive (which is what 'increasing' implies for r), it means that as the angle θ increases, the radius r also increases. An increasing radius means the point is moving farther away from the origin, or pole. Moving closer to the pole would mean r is decreasing (Choice A). The movement being counter-clockwise (Choice C) is the default way θ increases, but it doesn't depend on the rate of change of r.

To test if a function is even or odd, we evaluate f(-x). f(-x) = 2(-x)^3 - 8(-x) = 2(-x^3) + 8x = -2x^3 + 8x. Now we compare this to the original f(x). We can see that f(-x) = -(2x^3 - 8x) = -f(x). Since f(-x) = -f(x), the function is odd. An even function would have f(-x) = f(x). A function is odd if all its terms have odd powers of x (here, powers 3 and 1), and it's even if all its terms have even powers.

The distance from the pole is given by r. We want to maximize r = 3 + 3cos(θ). The value of r will be at its maximum when the cos(θ) term is at its maximum. The maximum value of cos(θ) is 1. This occurs when θ = 0 (and other coterminal angles like 2π, etc.). At θ=0, r = 3 + 3cos(0) = 3 + 3(1) = 6. At θ=π, cos(π)=-1, which gives the minimum distance r=0.

This is a logistic growth model. The long-term carrying capacity is the value that P(t) approaches as t goes to infinity. As t -> ∞, the term e^(-0.05t) approaches 0 (since it's 1 / e^(0.05t) and the denominator gets huge). So the equation becomes P(t) -> 120 / (1 + 3*0) = 120 / 1 = 120. The carrying capacity is the value in the numerator of the logistic function. The initial population at t=0 would be 120 / (1+3) = 30 (Choice A), which is a common distractor.

The roots (or zeros) of a polynomial are the values of x that make P(x) = 0. Since the polynomial is already factored, we can find the roots by setting each factor to zero. (x-3)^2 = 0 gives a root of x=3. (x+1) = 0 gives a root of x=-1. (x^2+4) = 0 gives x^2 = -4, so x = ±sqrt(-4) = ±2i. The roots are 3, -1, 2i, and -2i. The value x=2 is not a root.

This is a challenging question about function composition and rates of change. Let's analyze the first derivative (rate of change). By the chain rule, h'(x) = f'(g(x)) * g'(x). Since f is decreasing, f' is negative. Since g is increasing, g' is positive. So h'(x) = (negative) * (positive) = negative. This means h is decreasing. Now for concavity (the second derivative). h''(x) is more complex. A simpler way to think about it: as x increases, g(x) increases (because g is increasing). As the input to f increases, the output f(g(x)) decreases (because f is decreasing). So h is decreasing. For concavity: g is concave down, so its rate of increase is slowing. f is concave up, so its rate of decrease is slowing. As x increases, g(x) increases at a slower rate. f is applied to these slowing increasing values. Because f is concave up, its slope becomes less negative as its input increases. The combination of these effects results in h being concave up. The rate of decrease of h slows down.

This question connects the concepts of average and instantaneous rates of change. First, find the instantaneous rate of change at x=2. The formula is given: 3x^2. So at x=2, the rate is 3(2^2) = 12. Now we need to find which interval has an average rate of change of 12. Let's test the choices. For Choice C, [1, sqrt(7)]: The average rate of change is (f(sqrt(7)) - f(1)) / (sqrt(7) - 1) = ((sqrt(7))^3 - 1^3) / (sqrt(7) - 1) = (7*sqrt(7) - 1) / (sqrt(7) - 1). This looks complicated. Let's re-think. We want (b^3 - a^3) / (b-a) = 12. The numerator is a difference of cubes, which factors to (b-a)(b^2 + ab + a^2). So we want b^2 + ab + a^2 = 12. Let's test the choices with this formula. Choice A: 3^2 + 1*3 + 1^2 = 9+3+1 = 13. No. Choice B: (sqrt(12))^2 + 0*sqrt(12) + 0^2 = 12. Yes. Let me re-check my first calculation. Ah, I made a mistake. Let's re-evaluate choice B. (f(sqrt(12)) - f(0)) / (sqrt(12) - 0) = (sqrt(12))^3 / sqrt(12) = (sqrt(12))^2 = 12. So B is the correct answer. Let's re-evaluate C to be sure. (sqrt(7))^2 + 1*sqrt(7) + 1^2 = 7 + sqrt(7) + 1 = 8 + sqrt(7). No. The correct answer is B.

This function is a product of a linear function (5t) and an exponential decay function (e^(-0.5t)). Initially, for small t, the linear term 5t dominates, causing the concentration to increase from C(0)=0. However, as t gets large, the exponential decay term e^(-0.5t) dominates and goes to zero much faster than 5t goes to infinity. This will pull the entire function's value down towards zero. Therefore, the function must rise from zero to some maximum concentration and then fall back down, approaching zero in the long run. This is a very common shape for models of this type.

This requires using two Pythagorean identities. The numerator 1 - cos^2 x is equal to sin^2 x. For the denominator, we start with the identity 1 + cot^2 x = csc^2 x. Rearranging this gives cot^2 x = csc^2 x - 1. So our expression becomes sin^2 x / cot^2 x. Now, we convert cot^2 x to its definition in terms of sine and cosine: cot^2 x = cos^2 x / sin^2 x. The expression is now sin^2 x / (cos^2 x / sin^2 x). To divide by a fraction, we multiply by the reciprocal: sin^2 x * (sin^2 x / cos^2 x). This simplifies to sin^4 x / cos^2 x.

The average rate of change on an interval [a, b] is positive if f(b) > f(a). We need to test the endpoints for each interval. The function has roots at x=3 (a double root, so it touches the axis here) and x=-1. A) On [-1, 3], f(-1)=0 and f(3)=0. The average rate of change is 0. B) On [1, 3], f(1) = 2(1-3)^2(1+1) = 16 and f(3)=0. The average rate of change is (0-16)/(3-1) = -8, which is negative. C) On [3, 4], f(3)=0 and f(4) = 2(4-3)^2(4+1) = 10. The average rate of change is (10-0)/(4-3) = 10, which is positive. D) On [0, 2], f(0) = 2(0-3)^2(0+1) = 18 and f(2) = 2(2-3)^2(2+1) = 6. The average rate of change is (6-18)/(2-0) = -6, which is negative.

Let's find the parameters. The diameter is 100, so the radius (amplitude a) is 50. The center is 60 feet high, so the midline d is 60. The period is 8 minutes. The formula for the period is P = 2π/b, so 8 = 2π/b, which gives b = 2π/8 = π/4. The rider starts at the bottom. A standard cosine function starts at the top, and a standard sine function starts at the midline. To start at the bottom, we need a reflected cosine function, which means using -cos. Putting it all together: amplitude a=50, b=π/4, midline d=60, and a negative cosine shape. This gives h(t) = -50cos(π/4 * t) + 60.

This is a great shortcut question. The domain of the inverse function f⁻¹ is the same as the range of the original function f. The function f(x) = (x-2)/(x+3) is a rational function. To find its range, we can look for the horizontal asymptote. Since the degrees of the numerator and denominator are both 1, the horizontal asymptote is the ratio of the leading coefficients, which is y = 1/1 = 1. The function will get infinitely close to y=1 but never touch it. Therefore, the range of f is all real numbers except 1. This means the domain of f⁻¹ is also all real numbers except 1.

The point (8, 3/2) being on the graph means that when x=8, f(x)=3/2. So we can write the equation 3/2 = log_b(8). To solve for b, we convert this logarithmic equation into its exponential form: b^(3/2) = 8. To get b by itself, we can raise both sides to the reciprocal power, 2/3. So, (b^(3/2))^(2/3) = 8^(2/3). This simplifies to b = (8^(1/3))^2. The cube root of 8 is 2, so we have b = 2^2 = 4.

This is about how transformations affect coordinates. The transformation g(x) = 3f(x/2) - 1 has three parts. The x/2 inside the function affects the x-coordinate. This is a horizontal stretch by a factor of 2. So the new x-coordinate is 4 * 2 = 8. The 3 multiplying the function and the -1 outside affect the y-coordinate. It's a vertical stretch by a factor of 3, followed by a shift down by 1. So the new y-coordinate is 3 * 5 - 1 = 15 - 1 = 14. The new point is (8, 14). It's easy to get the horizontal transformation backwards and divide by 2 instead of multiplying (which would lead to Choice A).

This problem is all about exponent rules. In the numerator, when you multiply powers with the same base, you add the exponents: x^2 * x^(1/3) = x^(2 + 1/3). To add 2 + 1/3, we need a common denominator: 6/3 + 1/3 = 7/3. So the numerator is x^(7/3). Now the expression is x^(7/3) / x^(-1/2). When you divide powers with the same base, you subtract the exponents: x^(7/3 - (-1/2)) = x^(7/3 + 1/2). To add these fractions, we need a common denominator of 6: 14/6 + 3/6 = 17/6. So, a = 17/6.

This question tests the fundamental properties of different function families. For a linear function, the first differences are constant (this is the slope). For a quadratic function, the second differences are constant. For a cubic function, the third differences are constant. For an exponential function, the ratios of consecutive outputs are constant. Since the second differences are constant, a quadratic model is the most appropriate choice.

End behavior describes what happens to a function as x approaches positive or negative infinity. For periodic functions like tan(x), the values repeat in a cycle and never settle down to approach a single number or go to infinity in a predictable way. The tangent function oscillates between negative and positive infinity within each period. Therefore, we say the limit as x -> ∞ does not exist, and the end behavior cannot be determined in the same way as for a polynomial or exponential function.

This question asks about the behavior near a vertical asymptote. The denominator is zero when x=2. Let's consider values of x very close to 2. The term (x-2) will be a very small number (either positive or negative). When we square it, (x-2)^2 becomes a very small positive number. The function is 1 divided by a very small positive number. This results in a very large positive number. So, as x approaches 2 from either the left or the right, f(x) approaches positive infinity.

The amplitude of a sinusoidal function is half the difference between its maximum and minimum values. It represents the distance from the midline to a peak or a trough. In this case, the maximum is 15 hours and the minimum is 9 hours. The amplitude is (Max - Min) / 2 = (15 - 9) / 2 = 6 / 2 = 3. The midline, which represents the average number of daylight hours, would be (15 + 9) / 2 = 12 hours (Choice C is the midline, not the amplitude).

This question asks, 'To what power must we raise 4 to get 1/16?'. We know that 4^2 = 16. The negative exponent rule tells us that b^(-n) = 1/b^n. Therefore, 4^(-2) = 1/4^2 = 1/16. So, log_4(1/16) = -2.

A hole occurs when a factor in the denominator cancels with a factor in the numerator. A vertical asymptote occurs when a factor in the denominator does not cancel. For a hole at x=3, we need the factor (x-3) to be in both the top and bottom. For a vertical asymptote at x=-1, we need the factor (x+1) to be in the bottom but not the top (after cancellation). Choice A, f(x) = (x-3) / ((x-3)(x+1)), fits this perfectly. The (x-3) terms would cancel to create the hole, leaving 1/(x+1), which has the required vertical asymptote.