Question

let ( h(x)=x^{2}-6x ).
(a) find the average rate of change from 5 to 8.
(b) find an equation of the secant line containing ( (5, h(5)) ) and ( (8, h(8)) ).
(a) the average rate of change from 5 to 8 is ( square ). (simplify your answer.)
(b) an equation of the secant line containing ( (5, h(5)) ) and ( (8, h(8)) ) is ( square ).
(type your answer in slope - intercept form.)

Explanation:

Step1: Calculate \( h(5) \) and \( h(8) \)

For \( h(x)=x^{2}-6x \), when \( x = 5 \), \( h(5)=5^{2}-6\times5=25 - 30=-5 \).
When \( x = 8 \), \( h(8)=8^{2}-6\times8=64 - 48 = 16 \).

Step2: Find the average rate of change (for part (a))

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, \( a = 5 \), \( b = 8 \), \( h(5)=-5 \), \( h(8)=16 \).
So, the average rate of change is \( \frac{h(8)-h(5)}{8 - 5}=\frac{16-(-5)}{3}=\frac{16 + 5}{3}=\frac{21}{3}=7 \).

Step3: Find the equation of the secant line (for part (b))

The slope \( m \) of the line passing through \((x_1,y_1)=(5,-5)\) and \((x_2,y_2)=(8,16)\) is \( m = 7 \) (from part (a)).
Using the point - slope form \( y - y_1=m(x - x_1) \), with \( x_1 = 5 \), \( y_1=-5 \) and \( m = 7 \).
\( y-(-5)=7(x - 5) \)
\( y + 5=7x-35 \)
\( y=7x-40 \)

Answer:

(a) \( 7 \)
(b) \( y = 7x-40 \)