First, look at the zeroes of the graph. Zeroes are where the function touches the x-axis (i.e. values of $\displaystyle x$ where $\displaystyle f(x)=0$). 

The graph shows the function touching the x-axis when $\displaystyle x=-2$, $\displaystyle x=0$, and at a value in between 1.5 and 2.

Notice all of the possible answers are already factored. Therefore, look for one with a factor of $\displaystyle x$ (which will make $\displaystyle f(x)=0$ when $\displaystyle x=0$), a factor of $\displaystyle (x+2)$ to make $\displaystyle f(x)=0$ when $\displaystyle x=-2$, and a factor which will make $\displaystyle f(x)=0$ when $\displaystyle x$ is at a value between 1.5 and 2.

$\displaystyle f(x)= x(3x-5)(x+2)$

This function fills the criteria; it has an $\displaystyle x$ and an $\displaystyle (x+2)$ factor. Additionally, the third factor, $\displaystyle (3x-5)$, will result in $\displaystyle f(x)=0$ when $\displaystyle x=\frac{5}{3}$, which fits the image. It also does not have any extra zeroes that would contradict the graph.

Given two points 

$\displaystyle (x_{0},y_{0})$ and $\displaystyle (x_{1},y_{1})$

the formula for a slope is 

$\displaystyle m=\frac{y_{1}-y_{0}}{x_{1}-x_{0}}$.

Thus, since our given table is 

$\displaystyle \begin{tabular}{ | c | c |} \hline x & y \\ \hline 0 & 2 \\ \hline 1 & 3 \\ \hline 2 & 4 \\ \hline \end{tabular}$

we select two points, say

$\displaystyle (0,2)$ and $\displaystyle (1,3)$

and use the slope formula to compute the slop. 

Thus,

$\displaystyle m = \frac{3-2}{1-0}=\frac{1}{1} = 1$.

Hence, the slope of the line generated by the table is

$\displaystyle m=1$.

The average rate of change of a function $\displaystyle f$ over an interval $\displaystyle [a, b]$ is equal to

$\displaystyle \frac{f(b)-f(a)}{b-a}$

Setting $\displaystyle a = 2, b= 3$, this is

$\displaystyle \frac{f(3)-f(2)}{3-2} = \frac{f(3)-f(2)}{1} = f(3)-f(2)$

Evaluate $\displaystyle f(3)$ and $\displaystyle f(2)$ by substitution:

$\displaystyle f(x) = x^{2}+ x$

$\displaystyle f(3) = 3^{2}+ 3= 9+ 3 = 12$

$\displaystyle f(2) = 2^{2}+ 2 = 4+ 2 = 6$

$\displaystyle \frac{f(3)-f(2)}{3-2} = 12 - 6 = 6$,

the correct response.

The average rate of change of a function $\displaystyle f$ over an interval $\displaystyle [a, b]$ is equal to

$\displaystyle \frac{f(b)-f(a)}{b-a}$

Setting $\displaystyle a = -1, b= 1$, this is

$\displaystyle \frac{f(1)-f(-1)}{1-(-1)} = \frac{f(1)-f(-1)}{2}$

Evaluate $\displaystyle f(1)$ using the definition of $\displaystyle f$ for $\displaystyle x>0$:

$\displaystyle f(x) = x^{2}-x$

$\displaystyle f(1) = 1^{2}-1 = 1 - 1 = 0$

Evaluate $\displaystyle f(-1)$ using the definition of $\displaystyle f$ for $\displaystyle x\le 0$:

$\displaystyle f(x) = x^{2}+x$

$\displaystyle f(-1) = (-)1^{2}+ (-1) = 1 - 1 = 0$

The average rate of change is therefore

$\displaystyle \frac{f(1)-f(-1)}{2} = \frac{0-0}{2} = 0$.

The rate of change between two points on a curve can be approximated by calculating the change between two points. 

Let $\displaystyle (x_1,y_1)$ be the coordinates of the first point and $\displaystyle (x_2,y_2)$ be the coordinates of the second point. Then the formula giving approximate rate of change is:

$\displaystyle \frac{y_2 - y_1} {x_2 - x_1}$

Notice that the numerator is the overall change in y, and the denominator is the overall change in x.

The calculation for the problem proceeds as follows:

Let $\displaystyle (3,2)$ be the first point and $\displaystyle (8,5)$ be the second point. Substitute in the values from these coordinates: 

$\displaystyle \frac{y_2 - y_1} {x_2 - x_1} \rightarrow \frac {5-2}{8-3}$

Subtract to get the final answer:

$\displaystyle \frac{3}{5}$

Note that it does not matter which you assign to be the first point and which you assign to be the second, as it will lead to the same value due to negatives canceling out.

The rate of change of a function $\displaystyle f$ on the interval $\displaystyle [a, b]$ is equal to 

$\displaystyle r = \frac{f(b)-f(a)}{b - a}$.

Set $\displaystyle a = -2, b = -1.5$. Refer to the graph of the function below:

The graph passes through $\displaystyle (-2, 2)$ and $\displaystyle (-1.5, 5)$.

$\displaystyle f(-2) = 2 , f(-1.5)= 5$. Thus, 

$\displaystyle r = \frac{f(-1.5)-f(-2)}{-1.5 - (-2)} = \frac{5-2}{0.5} = \frac{3}{0.5} = 6$,

the correct response.

The rate of change of a function $\displaystyle f$ on the interval $\displaystyle [a, b]$ is equal to 

$\displaystyle r = \frac{f(b)-f(a)}{b - a}$.

Set $\displaystyle a = 1, b = 5, r =4$. Examine the figure below:

The graph passes through the point $\displaystyle (1, 2)$, so $\displaystyle f(1) = 2$. Therefore, 

$\displaystyle 4 = \frac{f(5)-f(1)}{5-1}$

and, substituting,

$\displaystyle 4 = \frac{f(5)-2}{4}$

Solve for $\displaystyle f(5)$ using algebra:

$\displaystyle \frac{f(5)-2}{4} \cdot 4 = 4 \cdot 4$

$\displaystyle f(5)-2 =16$

$\displaystyle f(5)-2+ 2 =16 + 2$

$\displaystyle f(5)= 18$,

the correct response.

The average rate of change of a function $\displaystyle f$ on the interval $\displaystyle (a, b)$ is equal to 

$\displaystyle r = \frac{f(b)-f(a)}{b - a}$.

Restated, it is the slope of the line that passes through $\displaystyle (a, f(a))$ and $\displaystyle (b, f(b))$.

To find the correct value of $\displaystyle c$ that answers this question, it suffices to examine the line with slope $\displaystyle -\frac{4}{5}$ through $\displaystyle (0, f(0))$ and find the point among those given that is closest to the line. This line falls 4 units for every 5 horizontal units, so the line looks like this:

The $\displaystyle x$-coordinate of the point of intersection is closer to 2 than to any other of the values in the other four choices. This makes 2 the correct choice.