Calculate and interpret rate of change over a specified interval - HiSET

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What is the slope of the line given by the following table?

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

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Given two points

and

the formula for a slope is

.

Thus, since our given table is

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

we select two points, say

and

and use the slope formula to compute the slop.

Thus,

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

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

.

Define $f(x) = $x^{2}$+ x$ .

Give the average rate of change of over the interval .

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The average rate of change of a function over an interval is equal to

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

Setting , this is

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

Evaluate and by substitution:

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

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

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

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

the correct response.

What is the slope of the line given by the following table?

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

Tap to see back →

Given two points

and

the formula for a slope is

.

Thus, since our given table is

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

we select two points, say

and

and use the slope formula to compute the slop.

Thus,

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

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

.

Define $f(x) = $x^{2}$+ x$ .

Give the average rate of change of over the interval .

Tap to see back →

The average rate of change of a function over an interval is equal to

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

Setting , this is

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

Evaluate and by substitution:

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

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

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

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

the correct response.

What is the slope of the line given by the following table?

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

Tap to see back →

Given two points

and

the formula for a slope is

.

Thus, since our given table is

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

we select two points, say

and

and use the slope formula to compute the slop.

Thus,

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

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

.

Define $f(x) = $x^{2}$+ x$ .

Give the average rate of change of over the interval .

Tap to see back →

The average rate of change of a function over an interval is equal to

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

Setting , this is

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

Evaluate and by substitution:

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

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

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

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

the correct response.

What is the slope of the line given by the following table?

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

Tap to see back →

Given two points

and

the formula for a slope is

.

Thus, since our given table is

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

we select two points, say

and

and use the slope formula to compute the slop.

Thus,

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

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

.

Define $f(x) = $x^{2}$+ x$ .

Give the average rate of change of over the interval .

Tap to see back →

The average rate of change of a function over an interval is equal to

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

Setting , this is

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

Evaluate and by substitution:

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

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

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

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

the correct response.

Define $f(x) = \left\{\begin{matrix} $x^{2}$+x ,&x \le 0 \ $x^{2}$-x, & x> 0\end{matrix}\right.$ .

Give the average rate of change of over the interval .

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The average rate of change of a function over an interval is equal to

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

Setting , this is

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

Evaluate using the definition of for :

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

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

Evaluate using the definition of for $x\le 0$ :

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

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

The average rate of change is therefore

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

Define $f(x) = \left\{\begin{matrix} $x^{2}$+x ,&x \le 0 \ $x^{2}$-x, & x> 0\end{matrix}\right.$ .

Give the average rate of change of over the interval .

Tap to see back →

The average rate of change of a function over an interval is equal to

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

Setting , this is

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

Evaluate using the definition of for :

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

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

Evaluate using the definition of for $x\le 0$ :

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

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

The average rate of change is therefore

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

Define $f(x) = \left\{\begin{matrix} $x^{2}$+x ,&x \le 0 \ $x^{2}$-x, & x> 0\end{matrix}\right.$ .

Give the average rate of change of over the interval .

Tap to see back →

The average rate of change of a function over an interval is equal to

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

Setting , this is

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

Evaluate using the definition of for :

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

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

Evaluate using the definition of for $x\le 0$ :

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

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

The average rate of change is therefore

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

Define $f(x) = \left\{\begin{matrix} $x^{2}$+x ,&x \le 0 \ $x^{2}$-x, & x> 0\end{matrix}\right.$ .

Give the average rate of change of over the interval .

Tap to see back →

The average rate of change of a function over an interval is equal to

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

Setting , this is

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

Evaluate using the definition of for :

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

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

Evaluate using the definition of for $x\le 0$ :

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

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

The average rate of change is therefore

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