How do you find the average rate of change of #y=2x^2-2x+1# over [-1,-1/2]?

Question

Elijah Abram · Accepted Answer

#-5#

The #color(blue)"average rate of change"# of y over an interval between 2 points (a ,f(a)) and (b,f(b)) is the slope of the #color(blue)"secant line"# connecting the 2 points.

in order to determine the average rate of change between the two usage points.

#color(red)(bar(ul(|color(white)(a/a)color(black)((f(b)-f(a))/(b-a))color(white)(a/a)|)))#

#f(b)=f(-1/2)=2(-1/2)^2-2(-1/2)+1=2 1/2#

#f(a)=f(-1)=2(-1)^2-2(-1)+1=5#

The average rate of change between #(-1,5)" and " (-1/2,2 1/2)# is

#(2 1/2-5)/(-1/2-(-1))=(-2 1/2)/(1/2)=-5#

This means that the average of all the slopes of lines tangent to the graph of y between #(-1,5)"and " (-1/2,2 1/2)# is - 5

Charles Arocha · Answer

undefined To find the average rate of change of the function $y = 2x^2 - 2x + 1$ over the interval $[-1, -\frac{1}{2}]$, you can use the formula:

$$
\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}
$$

where $f(x)$ is the function and $a$ and $b$ are the endpoints of the interval.

In this case, $a = -1$ and $b = -\frac{1}{2}$.

First, find $f(-1)$ and $f\left(-\frac{1}{2}\right)$ by substituting the respective values of $x$ into the function:

$$f(-1) = 2(-1)^2 - 2(-1) + 1 = 2 - (-2) + 1 = 5$$
$$f\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right)^2 - 2\left(-\frac{1}{2}\right) + 1 = 2\left(\frac{1}{4}\right) + 1 + 1 = \frac{5}{2}$$

Now, substitute these values into the formula for average rate of change:

$$
\text{Average Rate of Change} = \frac{\frac{5}{2} - 5}{-\frac{1}{2} - (-1)}
$$

$$
= \frac{\frac{5}{2} - 5}{-\frac{1}{2} + 1}
$$

$$
= \frac{\frac{5}{2} - \frac{10}{2}}{\frac{-1}{2} + \frac{2}{2}}
$$

$$
= \frac{-\frac{5}{2}}{\frac{1}{2}}
$$

$$
= -\frac{5}{2} \times 2
$$

$$
= -5
$$

Therefore, the average rate of change of $y = 2x^2 - 2x + 1$ over the interval $[-1, -\frac{1}{2}]$ is $-5$.