Mock Exam #2: AP Precalculus

Let's look at the changes in each function. For every increase of 1 in x, f(x) increases by a constant amount: 8-5=3, 11-8=3, and so on. This means f(x) is a linear function with a constant rate of change. Now look at g(x). For each increase of 1 in x, g(x) is multiplied by 2. The change itself is not constant: 2-1=1, 4-2=2, 8-4=4. The rate of change is increasing. This is characteristic of an exponential function. Therefore, f(x) has a constant rate of change and g(x) has an increasing rate of change.

The average rate of change is just the slope between the endpoints of the interval. The formula is (f(b) - f(a)) / (b - a). First, let's find the function values. f(3) = (3)³ - 2(3) + 1 = 27 - 6 + 1 = 22. And f(-1) = (-1)³ - 2(-1) + 1 = -1 + 2 + 1 = 2. Now, plug these into the formula: (22 - 2) / (3 - (-1)) = 20 / 4 = 5. A common mistake is to just calculate f(3) - f(-1) and forget to divide by the change in x. Another slip-up is to miscalculate the denominator, like 3-1=2.

End behavior is determined by the leading term of the polynomial—the term with the highest degree. The problem states that the ends of the graph go in opposite directions. This tells you the degree of the polynomial must be odd. This eliminates choices A and D. For the remaining choices, we look at the leading coefficient. As x gets very large and positive (x → ∞), we want p(x) to be negative. In choice C, x³ would be positive. In choice B, -x⁵ would be negative. So, p(x) = -x⁵ + 3x² - 1 fits the description perfectly. It has an odd degree (5) and a negative leading coefficient (-1).

This question tests the Complex Conjugate Root Theorem. The theorem states that if a polynomial has real coefficients, complex roots must come in conjugate pairs. You are given a complex root: 2 - 3i. This means its conjugate, 2 + 3i, MUST also be a root. You are also given a real root, 4. So, the minimum set of roots we know are 4, (2 - 3i), and (2 + 3i). That's a total of three roots. A polynomial with 3 roots must have a minimum degree of 3. The most common mistake is to forget that the complex root implies its conjugate is also a root, leading to an answer of 2.

The first step for any rational function problem is to factor everything you can. The numerator, x² - 9, is a difference of squares: (x - 3)(x + 3). The denominator, x² + x - 12, factors to (x + 4)(x - 3). So, f(x) = [(x - 3)(x + 3)] / [(x + 4)(x - 3)]. Now, look for common factors that can be canceled. The (x - 3) term appears in both the top and bottom. This is where the hole is! Set that factor to zero: x - 3 = 0, so there's a hole at x = 3. The factor that's left in the denominator, (x + 4), causes a vertical asymptote. Set that factor to zero: x + 4 = 0, so the vertical asymptote is at x = -4. This is where students often slip up: you must factor and cancel before you declare your asymptotes.

Let's break down the transformations step-by-step. We start with f(x). The transformation f(x - 2) is a horizontal shift. Remember, it's a little counter-intuitive: a minus sign inside the parentheses means a shift to the RIGHT. So, f(x - 2) shifts the entire graph 2 units to the right. The vertex moves from (0,0) to (2,0). Next, we apply the negative sign out front, -f(x - 2). This is a reflection across the x-axis. The V-shape, which was opening upwards, will now open downwards. The vertex at (2,0) stays where it is because it's on the axis of reflection. So, the final graph is a V-shape opening down with its vertex at (2, 0).

This is an exponential growth model, which follows the form A(b)^t. The 'A' value is the initial amount, and 'b' is the growth factor. In our function, P(t) = 1200(1.04)^t, the initial amount is 1200. Since t is years since 2010, this was the population in 2010. The growth factor is 1.04. To find the growth rate, we use the formula b = 1 + r. So, 1.04 = 1 + r, which means r = 0.04. As a percentage, that's 4%. So, the population was 1200 in 2010 and it grows by 4% each year. A common mistake is to see 1.04 and think it's a 104% growth rate, but the rate is the part added to 1.

Function composition means we are plugging one function into another. The notation f(g(x)) tells you to take the entire function g(x) and substitute it for the 'x' in the function f(x). So, we start with f(x) = √(x - 3). Now, instead of x, we're going to write g(x), which is x² + 1. This gives us f(g(x)) = √((x² + 1) - 3). Simplifying inside the square root gives us √(x² - 2). Be careful not to get the order wrong; g(f(x)) would be a completely different function.

The core concept of an inverse function is that it 'undoes' the original function. If f(x) takes an input of 2 and gives an output of 8, then the inverse function, f⁻¹(x), must take an input of 8 and give an output of 2. In terms of coordinates, this means you simply swap the x and y values. So, if the point (2, 8) is on the graph of f(x), then the point (8, 2) must be on the graph of f⁻¹(x). It's a direct and simple rule.

This is all about using your logarithm properties. First, use the power rule: 2log₃(x) becomes log₃(x²). Now our expression is log₃(x²) + log₃(y) - log₃(z). Next, use the product rule for the terms being added: log₃(x²) + log₃(y) becomes log₃(x²y). Our expression is now log₃(x²y) - log₃(z). Finally, use the quotient rule for the subtraction: this becomes log₃(x²y / z). The key is to apply the rules one at a time. A common error is to misapply the power rule and write 2log₃(x) as log₃(2x).

To solve an exponential equation where the variable is in the exponent, you need to use logarithms. Since the base is 'e', the natural logarithm (ln) is the perfect tool. Take the natural log of both sides: ln(e^(2x)) = ln(9). One of the key properties of logs is that ln(e^stuff) = stuff. So, the left side simplifies to just 2x. Our equation is now 2x = ln(9). To get x by itself, divide both sides by 2: x = ln(9) / 2. A common mistake is to think that ln(9)/2 is the same as ln(4.5), but that's not a valid log property. Another mistake is to think e^(2x)=9 is the same as (e^x)^2=9, so e^x=3, which gives x=ln(3). This is also correct since ln(9)/2 = ln(3^2)/2 = 2ln(3)/2 = ln(3). However, choice A is the most direct result of the standard algorithm, and choice D is a simplified form. In AP questions, both would be considered correct, but let's check the options. Ah, I see both ln(9)/2 and ln(3) are options. Let's re-evaluate. ln(9) ≈ 2.197. So x ≈ 1.0985. ln(3) ≈ 1.0986. They are indeed the same value. Let's stick with A as the most direct algebraic step, but recognize D is an equivalent, simplified form. On the AP exam, you'd likely only see one of these correct forms.

The entire purpose of a semi-log plot is to transform an exponential relationship into a linear one. If your original data follows y = a b^x, and you take the log of both sides, you get log(y) = log(a) + x log(b). If you think of 'log(y)' as your new vertical axis, this equation is in the form Y = C + mX, which is the equation of a line. The slope of this line will be log(b). As long as the base 'b' is not 1, the slope will be non-zero. So, exponential data always appears as a straight, sloped line on a semi-log plot. This is a powerful tool for identifying exponential models.

First, locate the angle 7π/6 on the unit circle. Since 6π/6 is π (or 180°), 7π/6 is just a little more than that, placing it in Quadrant III. In Quadrant III, both the x-coordinate (cosine) and the y-coordinate (sine) are negative. The reference angle for 7π/6 is π/6. We know that cos(π/6) = √3/2. Since our angle is in Quadrant III where cosine is negative, the answer must be -√3/2. A common mistake is to forget the sign and choose √3/2, or to mix up the cosine and sine values and choose -1/2.

The period of a sinusoidal function of the form y = Acos(Bx - C) + D is determined by the 'B' value, the coefficient of x. The formula is Period = 2π / |B|. In this function, g(x) = 8cos(πx - 3) + 5, the B value is π. So, the period is 2π / π = 2. Don't get distracted by the other numbers! The 8 is the amplitude, the 3 affects the phase shift, and the 5 is the vertical shift. None of them change the period.

Let's build the model piece by piece. The midline (vertical shift) is the average of the high and low tides: D = (14 + 2) / 2 = 8. The amplitude is the distance from the midline to the high or low: A = 14 - 8 = 6. The time from a high to a low is 10 AM - 4 AM = 6 hours. This is half a period. So, the full period is 12 hours. The period is related to B by Period = 2π/B, so 12 = 2π/B, which gives B = 2π/12 = π/6. Since the function starts at a high point at t=4, it's easiest to use a cosine function with a phase shift of 4 to the right. Putting it all together: h(t) = 6cos(π/6 * (t - 4)) + 8.

This question asks for the angle whose sine is -√2/2. The critical thing to remember is that the arcsin function has a restricted range: [-π/2, π/2], which is Quadrant I and Quadrant IV. We need a negative sine value, so we must be in Quadrant IV. The reference angle where sin(θ) = √2/2 is π/4. The angle in Quadrant IV that corresponds to this reference angle and is within our restricted range is -π/4. Choices A and B, 7π/4 and 5π/4, both have a sine of -√2/2, but they are outside the required range for the arcsin function. This is a very common trap that students fall into.

We are looking for angles on the unit circle where tan(x) = 1. Remember that tan(x) = sin(x) / cos(x). For the tangent to be 1, the sine and cosine must be equal. This happens in Quadrant I, where both are positive, and in Quadrant III, where both are negative. The angle in Quadrant I is x = π/4. The angle in Quadrant III is x = 5π/4. Both of these values are within the interval [0, 2π). Therefore, there are exactly 2 solutions.

Let's visualize the point (0, -5) in the Cartesian plane. It's on the y-axis, 5 units down from the origin. In polar coordinates, 'r' is the distance from the origin, and 'θ' is the angle from the positive x-axis. The distance from the origin to (0, -5) is clearly 5, so r=5. Now for the angle. We start at the positive x-axis (0 radians) and rotate counter-clockwise. To get to the negative y-axis, we need to rotate through Quadrant I and II (which is π radians) and then another half-quadrant (π/2 radians). The total angle is π + π/2 = 3π/2. So the polar coordinates are (5, 3π/2). Choice C, (-5, π/2), is also a valid representation, but it's not among the primary positive-r choices and B is a more direct representation.

The vertex of this upward-opening parabola is at x=3. This is the minimum point of the function. This means the function is decreasing for all x < 3 and increasing for all x > 3. The average rate of change will be positive over any interval where the function is consistently increasing. Let's look at the choices. [0, 3] is an interval where the function is decreasing. [1, 5] and [2, 4] both cross over the vertex at x=3, so the function decreases and then increases, making the sign of the average rate of change uncertain without the full function. However, the interval [4, 7] is entirely to the right of the vertex (since 4 > 3). Over this entire interval, the function is increasing, which guarantees that the average rate of change will be positive.

This is a classic real-world example of exponential decay, but with a twist. The temperature doesn't decay to zero, it decays towards the room temperature. The difference between the coffee's temperature and the room's temperature is what follows an exponential decay model. The function would look something like T(t) = T_room + (T_initial - T_room)e^(-kt). This is a transformed exponential function. Among the choices given, 'Exponential' is the family of functions that captures this behavior of changing by a certain percentage of the remaining difference over time. A linear function would mean it cools by the same number of degrees every minute, which isn't realistic. A quadratic function would imply it starts cooling down and then heats back up, which makes no sense.

Alright, let's break this one down. First, we use the logarithm product rule to combine the two logs: log₂(x(x - 7)) = 3. Now, we convert this from logarithmic to exponential form: x(x - 7) = 2³. This simplifies to x² - 7x = 8, or x² - 7x - 8 = 0. Factoring this quadratic gives us (x - 8)(x + 1) = 0, which means potential solutions are x = 8 and x = -1. Here's the critical step many students miss: you must check for extraneous solutions. The domain of log(a) requires 'a' to be positive. If we plug x = -1 back into the original equation, we get log₂(-1), which is undefined. So, x = -1 is an extraneous solution. Only x = 8 is valid. Choice B is tempting but forgets to check the domain. Choice C comes from incorrectly adding the arguments of the logs (log₂(x + x - 7)). Choice D comes from a calculation error, like thinking 2³ is 6 instead of 8.

This question asks for the average rate of change, so you should immediately think of the slope formula: (y₂ - y₁) / (x₂ - x₁). In this polar context, it's (r(θ₂) - r(θ₁)) / (θ₂ - θ₁). Let's calculate the values we need. First, r(π) = 3 + 2cos(π) = 3 + 2(-1) = 1. Next, r(π/2) = 3 + 2cos(π/2) = 3 + 2(0) = 3. Now, plug these into the formula: (1 - 3) / (π - π/2) = -2 / (π/2). To divide by a fraction, we multiply by its reciprocal: -2 * (2/π) = -4/π. Choice B is a common sign error. Choice C is what you get if you correctly find the change in r, which is -2, but forget to divide by the change in θ. Choice D happens if you divide by π instead of the interval length, π/2.

To figure out end behavior, you only need to look at the leading term of the polynomial. You can find this by multiplying the highest power of x from each factor. We have -(x)², (x), and (x). Multiplying them together gives -(x²)(x)(x) = -x⁴. The degree of the polynomial is 4 (an even number) and the leading coefficient is -1 (a negative number). For any even-degree polynomial with a negative leading coefficient, both 'arms' of the graph point down. This means as x approaches both positive and negative infinity, p(x) approaches negative infinity. So, choice B is our answer. Choice A describes a polynomial with an odd degree and negative leading coefficient. Choice D describes one with an odd degree and positive leading coefficient. Choice C is for an even degree with a positive leading coefficient.

Function composition can look tricky, but just work from the inside out. We need to find g(f(0)). The first step is to find the value of the inner function, f(0). Look at the table in the row where x = 0. The value of f(x) in that row is 1. So, f(0) = 1. Now we can substitute this value back into our expression: we are looking for g(1). Go back to the table and find the row where x = 1. In that row, the value of g(x) is 0. Therefore, g(f(0)) = 0. The most common mistake is to calculate f(g(0)) instead. That would be f(-2), which equals 0. In this case, you'd get the right answer by accident, but the process is wrong! Another mistake is to multiply values, which isn't what composition means.

Let's build the equation piece by piece from the graph. First, the midline is the average of the max and min y-values: (2 + (-4))/2 = -1. This is our vertical shift, D = -1. The amplitude is the distance from the midline to a max or min: |2 - (-1)| = 3. So, A = 3. The period is the horizontal distance for one full cycle. The graph goes from a maximum at x=0 to the next maximum at x=π, so the period is π. We find B using the formula Period = 2π/B. So, π = 2π/B, which means B = 2. Since the graph starts at a maximum at x=0, it's a standard (not reflected, not phase-shifted) cosine function. Putting it all together: y = Acos(Bx) + D gives us y = 3cos(2x) - 1. Choice C would be correct if the graph started at a minimum. Choice B comes from an error in calculating the period. Choice D incorrectly uses the total vertical distance (max to min) as the amplitude.

This is a classic question that tests if you can tell the difference between a vertical asymptote and a hole. The first step is always to factor the numerator and the denominator. The numerator becomes (x - 2)(x + 2). The denominator factors to (x - 3)(x + 2). So, our function is f(x) = [(x - 2)(x + 2)] / [(x - 3)(x + 2)]. Notice the (x + 2) factor appears in both the top and bottom. This means we can cancel it out, which creates a hole in the graph at x = -2. A vertical asymptote occurs where the denominator is zero after canceling. The remaining factor in the denominator is (x - 3). Setting this to zero gives x = 3. This is our vertical asymptote. Choice B is the most common mistake: it lists all the values that make the original denominator zero, without checking for holes. Choice C incorrectly identifies the hole as an asymptote. Choice D gives the horizontal asymptote, not the vertical one.

This is a great conceptual question. The key is understanding what a semi-log plot does. A semi-log plot uses a logarithmic scale on one axis (in this case, the y-axis for population P). If a plot of (t, log(P)) is linear, it means that log(P) is a linear function of t. We can write this as log(P) = mt + c. To find the relationship for P, we need to undo the logarithm. We do this by making both sides the exponent of the base of the logarithm (let's assume base 10). This gives 10^(log(P)) = 10^(mt + c). This simplifies to P = 10^(mt) 10^c. We can rewrite this as P = (10^c) (10^m)^t. If we let a = 10^c and b = 10^m, our equation is P = a * b^t, which is the form of an exponential function. Choice A is incorrect because the original plot was a curve. Choice C would be true if a plot of (log(t), P) was linear. Choice D would be true if a log-log plot, (log(t), log(P)), was linear.

Let's find this value on the unit circle. The angle 5π/6 is in the second quadrant (since it's slightly less than π, which is 6π/6). The reference angle is the acute angle it makes with the x-axis, which is π - 5π/6 = π/6. We know that cos(π/6) = √3/2. Now, we just need the correct sign. In the second quadrant, x-coordinates are negative, and cosine corresponds to the x-coordinate on the unit circle. Therefore, cos(5π/6) must be negative. So, cos(5π/6) = -√3/2. Choice A has the right value but the wrong sign. Choice C is the value of sin(5π/6) or cos(2π/3), a common mix-up.

Let's draw a picture. We have a rectangle with one side against a wall. Let the side parallel to the wall be 'x'. Let the other two fenced sides (perpendicular to the wall) each have length 'y'. The area is length times width, so x * y = 200. The problem asks for a function in terms of x, so we should solve this for y: y = 200/x. Now, let's think about the perimeter, which is the length of the fence. We need one piece of length 'x' and two pieces of length 'y'. So, the total fence is P = x + 2y. Now, substitute our expression for y: P(x) = x + 2(200/x), which simplifies to P(x) = x + 400/x. Choice B is what you'd get if you fenced all four sides (2x + 2y). Choice C is a common error where you forget to multiply by 2 for the two sides of length y. Choice D incorrectly sets up the relationship between the variables.

To find the inverse of a function, we start by writing it as y = (2x + 1) / (x - 3). Then, we swap the x and y variables: x = (2y + 1) / (y - 3). Now, the hard part: we have to solve this equation for y. First, multiply both sides by (y - 3) to get x(y - 3) = 2y + 1. Distribute the x: xy - 3x = 2y + 1. The trick here is to get all the terms with y on one side and everything else on the other. So, xy - 2y = 3x + 1. Now, factor out the y: y(x - 2) = 3x + 1. Finally, divide to isolate y: y = (3x + 1) / (x - 2). So, f⁻¹(x) = (3x + 1) / (x - 2). Choice A is a common guess, just flipping the fraction, but that's not how inverses work. Choices C and D are results of sign errors during the algebraic manipulation. Note that choice C is actually equivalent to B, but B is the standard simplified form.

This equation looks like a quadratic equation, but with sin(θ) instead of x. Let's make a substitution: let u = sin(θ). The equation becomes 2u² - u - 1 = 0. We can factor this into (2u + 1)(u - 1) = 0. This gives us two possibilities: 2u + 1 = 0 (so u = -1/2) or u - 1 = 0 (so u = 1). Now, we substitute back sin(θ) for u. Case 1: sin(θ) = 1. On the interval [0, 2π), this happens only at θ = π/2. Case 2: sin(θ) = -1/2. Sine is negative in Quadrants III and IV. The reference angle where sin is 1/2 is π/6. So, the angles are π + π/6 = 7π/6 (in Q3) and 2π - π/6 = 11π/6 (in Q4). Combining all our solutions, we get {π/2, 7π/6, 11π/6}. Choice C gets the sign wrong for sin(θ) = 1 and uses the wrong quadrants for sin(θ) = -1/2. Choice B makes a reference angle error for sin(θ) = -1/2.

This expression is a complex fraction, which often looks more intimidating than it is. Let's tackle the numerator first. (x+h)⁻¹ is 1/(x+h), and x⁻¹ is 1/x. So the numerator is (1/(x+h)) - (1/x). To combine these, we need a common denominator, which is x(x+h). This gives us [x - (x+h)] / [x(x+h)]. Be careful with the parentheses! The numerator simplifies to x - x - h = -h. So our whole expression is now ( -h / [x(x+h)] ) / h. Dividing by h is the same as multiplying by 1/h. So we have ( -h / [x(x+h)] ) * (1/h). The h in the numerator and the h in the denominator cancel out, leaving us with -1 / (x(x+h)). This is a very important expression in calculus, called the difference quotient for the function f(x)=1/x. Choice B has a sign error. Choice C is what you get if you incorrectly assume h is zero before simplifying, which isn't allowed.

Let's evaluate each part of the expression separately. First, log₃(27). This asks the question: '3 to what power equals 27?' Since 3³ = 27, we know log₃(27) = 3. Next, log₂(1/4). This asks: '2 to what power equals 1/4?' Since 2² = 4, we know 2⁻² = 1/4. So, log₂(1/4) = -2. Now we just substitute these values back into the original expression: 3 - (-2). Subtracting a negative is the same as adding a positive, so 3 + 2 = 5. A common mistake is in the second term, leading to 3 - 2 = 1 (Choice A). Be careful with those negative exponents!

This composition of functions requires us to build a triangle. Let θ = arctan(-4/3). This means tan(θ) = -4/3. The range of the arctan function is (-π/2, π/2). Since tan(θ) is negative, θ must be in Quadrant IV. Now, let's draw a right triangle in Q4. We know tan(θ) = y/x. So we can set y = -4 and x = 3. (Remember, x is positive and y is negative in Q4). Now we find the hypotenuse, r, using the Pythagorean theorem: r = √(x² + y²) = √(3² + (-4)²) = √(9 + 16) = √25 = 5. The hypotenuse is always positive. The original question asks for cos(θ). We know cos(θ) = x/r. Using our values, cos(θ) = 3/5. Choice B is a common error from putting the triangle in the wrong quadrant (like Q2). Choice C would be sin(θ).

The average rate of change is just a fancy term for the slope between two points. The formula is (f(b) - f(a)) / (b - a). Here, our interval is [1, 3], so a=1 and b=3. From the table, we can see that f(3) = 15 and f(1) = 3. Now we plug these values into the formula: (15 - 3) / (3 - 1). This simplifies to 12 / 2, which equals 6. So, the average rate of change is 6. Be careful! A common mistake, Choice D, is to calculate the change in f(x), which is 12, but forget to divide by the change in x, which is 2. Choice C is what you get if you average the two y-values, (15+3)/2, which is not the rate of change.

Half-life problems are all about figuring out how many half-life periods have passed. The total time is 45 years, and one half-life is 15 years. So, the number of half-lives is 45 / 15 = 3. This means the amount of the substance will be cut in half three times. Let's track it: Start: 100 mg. After 1st half-life (15 years): 100 / 2 = 50 mg. After 2nd half-life (30 years): 50 / 2 = 25 mg. After 3rd half-life (45 years): 25 / 2 = 12.5 mg. So, after 45 years, 12.5 mg will remain. You can also use the formula A(t) = A₀ (1/2)^(t/h), where A₀=100, t=45, and h=15. This gives A(45) = 100 (1/2)^(45/15) = 100 (1/2)³ = 100 (1/8) = 12.5.

To convert from polar to rectangular coordinates, we use the formulas x = r cos(θ) and y = r sin(θ). Here, r = -2 and θ = 5π/3. First, let's find the values for the trig functions. The angle 5π/3 is in Quadrant IV. We know cos(5π/3) = 1/2 and sin(5π/3) = -√3/2. Now, let's calculate x and y. x = (-2) cos(5π/3) = (-2) (1/2) = -1. y = (-2) sin(5π/3) = (-2) (-√3/2) = √3. So the rectangular coordinates are (-1, √3). The negative 'r' value is what makes this tricky. It reflects the point (2, 5π/3), which is in Q4, through the origin, landing it in Q2. Our answer (-1, √3) is indeed in Q2, which is a good sanity check. Choice C is a common mistake where the negative sign on r is ignored. Choice B makes a sign error in the y-calculation.

This is a great test of your understanding of function transformations. Let's look at the two changes. The '+2' is inside the parentheses with the x. This indicates a horizontal shift. Here's where many students slip up: a positive value here means a shift to the LEFT. Think of it as needing a smaller x-value to get the same output as before. So, 'x+2' means shift left 2 units. The '-3' is outside the function, which indicates a vertical shift. This one is more straightforward: a negative value means a shift DOWN. So, '-3' means shift down 3 units. Combining these, the graph is shifted left 2 units and down 3 units. Choice C gets the horizontal shift direction wrong, which is the most common error.

This question is about correctly setting up the models for two different types of compound interest. For Option A, the keywords are 'compounded continuously'. This tells us to use the formula A = Pe^(rt). Here, P=$1000 and r=0.06, so the value is 1000e^(0.06t). For Option B, the keyword is 'compounded annually'. This means we use the formula A = P(1+r)^t. Here, P=$1200 and r=0.05, so the value is 1200(1+0.05)^t, or 1200(1.05)^t. The question asks when Option A's value exceeds Option B's value, which translates to the inequality: Value of A > Value of B. Plugging in our expressions gives 1000e^(0.06t) > 1200(1.05)^t. Choice B incorrectly uses the annual compounding formula for Option A. Choice C incorrectly uses the continuous compounding formula for Option B. Choice D uses a simple interest formula, which is incorrect for this problem.

The period of a periodic function is the length of one full cycle. The easiest way to measure this on a graph is to find the horizontal distance between two consecutive corresponding points, like two peaks or two troughs. The problem states that there is a peak at x = 1 and the very next peak is at x = 5. To find the distance between them, we just subtract the x-values: 5 - 1 = 4. So, the period of the function is 4. Don't be fooled by other features of the graph! The amplitude or the y-values of the peaks don't affect the period. Choice D is a common mistake where a student might just read the x-coordinate of the second peak instead of measuring the distance between the two.