We are told that f(x) must pass through the points (–1,5), (1,–1), and (2,–16). In order for it to pass through (–1,5), when x = –1 is input into the equation of f(x), the result must be 5. Likewise, when 1 and 2 are input into the equation of f(x), the results should be –1 and –16, respectively. To check if the equation for f(x) is correct, we need to put x = –1, 1, and 2 into the equation and see if we get the correct results. In short, we need to find the equation of f(x) such that f(–1) = 5, f(1) = –1, and f(2) = –16.

The only equation for f(x) that will pass through all three points is f(x) = 2x² – 3x³ . We can see this by testing x = –1, 1, and 2.

f(x) = 2x² – 3x³

f(–1) = 2(–1)² – 3(–1)³ = 2(1) – 3(–1) = 5

This means that (–1,5) would indeed lie on the graph of f(x).

f(1) = 2(1)² – 3(1)³ = 2 – 3 = –1

So f(x) would also pass through (1,–1).

f(2) = 2(2)² – 3(2)³ = 8 – 3(8) = –16

f(x) would also pass through (2,16).

We can check to see why the other equations of f(x) wouldn't satisfy all three points.

Let's look at the equation f(x) = –3x + 2 and test x = 2.

f(2) = –3(2) + 2 = –4, which doesn't equal –16. So that means this equation wouldn't pass through (2,–16).

Next, we can try f(x) = –1.5x² – 3x + 3.5, using x = 2.

f(2) = –1.5(2²) – 3(2) + 3.5 = –8.5, which is not –16.

Next, we can try f(x) = 1.5x² – 3x + 0.5 at x = 2.

f(2) = 1.5(2²) – 3(2) + 0.5 = 0.5, which is not equal to –16.

Finally, let's try f(x) = –x³ – 4x, using x = 1.

f(1) = –(1³) – 4(1) = –5, which doesn't equal –1.

The answer is f(x) = 2x² – 3x³ .

Let's go through each statement and determine if it's true.

f(x) = ax²

f(–x) = a(–x)² = a(–x)(–x) = ax² = f(x)

Statement I is true. Now let's look at the next statement.

f(x) = ax²

|f(x)| = |ax²|

Because a is positive, and because x² ≥ 0, the product of a and x² must be greater than or equal to zero. In other words, ax² ≥ 0. If we take the absolute value of a quantity that is greater than or equal to zero, we get that quantity. In general, if b ≥ 0, then |b| = b. This means that |ax²| = ax². Thus, we can write the following:

|f(x)| = |ax²| = ax² = f(x)

Therefore, statement II is also true. Let's look at the last statement.

f(f(x)) = f(ax²)= a(ax²)² = a(ax²)(ax²) = a³x⁴

This means III is also true.

The answer is I, II, and III.

A relation is a function if every x-value corresponds to one and only one y-value. Let's look at our answer choices. 

{(–3, –6), (–2, –1), (–1, 0), (0, 3), (1, 15)}: This is a function. There is only one y for each x-value.

{(–3, –6), (–2, –6), (–1, –6), (0, –6), (1, –6)}: This might not look like a fuction, but it is indeed a function. This set follows the rule that every x maps to only one y-value, it just happens that the y-value (-6) is the same for each x-value. 

{(–3, –6), (–2, –1), (–1, 0), (0, 3), (1, 0), (1, 15)}: This is NOT a function and therefore the correct answer. There are two x-values (both 1) that map to two different y-values, (1, 0) and (1, 15). Then this is a relation but NOT a function.

2y + 3x = 6: This is also a function. If you can solve for y, then the equation is a function. Here we can isolate y: y = –3x/2 + 3. 

y = 3x + 1: This is similar to the last answer choice, but with even less work to decide this is a function. Clearly this equation is in the form of "y = " and solves for a unique y-value for every x.

In the composite function g(f(x)), f(x) acts like "x" or any other variable. We need to plug f(x) into g(x). So g(f(x)) = 6(3x² + 4) – 1 = 18x² + 24 – 1 = 18x² + 23.

The function f(x) is a piecewise function, which means that it is comprised of separate functions that change depending on the value of x. According to the problem, whenever x is less than 1, f(x) is defined by 2x - 3. Whenever x is greater than or equal to one and less than 3, f(x) is defined as . And whenever x is greater than or equal to 3, f(x) is equal to |4x – 9|.

The question asks us to find f(1) + f(3) + f(5). We need to calculate the values of f(1), f(3), and f(5) individually and then find their sum.

To find f(1), we must first decide which of the three possible functions for f(x) to use. Since f(1) means we are finding the value of f(x) when x = 1, we will have to use the second piece, which says that f(x) = . We can't use the function 2x - 3, because this is only valid when x < 1, not when x = 1.

Next, we will find f(3). We need to use the third piece of the function which states that f(x) = |4x – 9|. Since f(3) means we are finding f(x) when x = 3, we can only use the third piece. The second piece, , is only defined if x is less than 3 and greater than or equal to 1.

Lastly, we must find f(5). Again, we will use the function f(x) = |4x – 9|, because when x = 5, x must be greater than 3.

We can now add up f(1), f(3), and f(5), which would give us 3 + 3 + 11 = 17.

The answer is 17.

First, let's expand . We will get the polynomial .

When we plug in the given function, g, into this polynomial and simplify, we will get .

Because this problem asks us for the value when x = 2, we have to plug in 2 for every x in the expression. .

This problem states that the given line crosses the y-axis at a certain point (m, n). By crossing the y-axis, we know that this point must be the y-intercept of this line. All y-intercepts lie on the y-axis and, therefore, have an x-coordinate of 0. If we plug in 0 for x in the equation of the line, , we will get

y = –6

The y-intercept must be at the point

(0, –6)

so n must be –6.

Plugging in  for  and 7 for  into  and solving for , we obtain .

Solving for  yields

.

We are solving for a composite function by substituting  into  to get: before simplification.