Question

let ( h(x)=x^{2}-7x ).
(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 ( square ). (simplify your answer.)
(b) an equation of the secant line containing ( (5, h(5)) ) and ( (7, h(7)) ) is ( square ).
(type your answer in slope - intercept form.)

Explanation:

Step1: Calculate \( h(5) \) and \( h(7) \)

For \( h(x)=x^{2}-7x \), when \( x = 5 \), \( h(5)=5^{2}-7\times5=25 - 35=-10 \).
When \( x = 7 \), \( h(7)=7^{2}-7\times7=49 - 49 = 0 \).

Step2: Find 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 \( a = 5 \), \( b = 7 \), \( h(5)=-10 \), \( h(7)=0 \).
So the average rate of change is \( \frac{h(7)-h(5)}{7 - 5}=\frac{0-(-10)}{2}=\frac{10}{2}=5 \).

Step3: Find the equation of the secant line

The slope \( m \) of the secant line is the average rate of change, so \( m = 5 \).
Using the point - slope form \( y - y_{1}=m(x - x_{1}) \), take the point \( (x_{1},y_{1})=(5,h(5))=(5,-10) \).
\( y-(-10)=5(x - 5) \).
Expand: \( y + 10=5x-25 \).
Solve for \( y \): \( y=5x-35 \).

Answer:

(a) \( 5 \)
(b) \( y = 5x-35 \)