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Question
let h(x)=x² - 2x. (a) find the average rate of change from 5 to 7. (b) find an equation of the secant line containing (5, h(5)) and (7, h(7)). (a) the average rate of change from 5 to 7 is 4 (simplify your answer.) (b) an equation of the secant line containing (5, h(5)) and (7, h(7)) is -4x - 35 (type your answer in slope - intercept form.)
Step1: Calculate \( h(5) \) and \( h(7) \)
Given \( h(x)=x^{2}-2x \).
For \( x = 5 \), \( h(5)=5^{2}-2\times5=25 - 10=15 \).
For \( x = 7 \), \( h(7)=7^{2}-2\times7=49 - 14 = 35 \).
Step2: Calculate the average rate of change
The formula for the average rate of change of a function \( y = h(x) \) from \( x=a \) to \( x = b \) is \( \frac{h(b)-h(a)}{b - a} \).
Here \( a = 5 \), \( b = 7 \), \( h(5)=15 \), \( h(7)=35 \).
So the average rate of change is \( \frac{h(7)-h(5)}{7 - 5}=\frac{35 - 15}{2}=\frac{20}{2}=10 \).
Step3: Find the equation of the secant line
The slope - intercept form of a line is \( y=mx + c \), where \( m \) is the slope and \( c \) is the \( y \) - intercept.
We know the slope \( m = 10 \) (from the average rate of change). Using the point - slope form \( y - y_{1}=m(x - x_{1}) \) with the point \( (x_{1},y_{1})=(5,15) \) (since \( h(5) = 15 \)).
\( y-15=10(x - 5) \).
Expand it: \( y-15=10x-50 \).
Add 15 to both sides: \( y = 10x-35 \).
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(a) The average rate of change from 5 to 7 is \( 10 \).
(b) The equation of the secant line is \( y = 10x-35 \).