Question

let ( h(x)=x^{2}-2x ).
(a) find the average rate of change from 4 to 6.
(b) find an equation of the secant line containing ( (4, h(4)) ) and ( (6, h(6)) ).
(a) the average rate of change from 4 to 6 is ( square ). (simplify your answer.)
(b) an equation of the secant line containing ( (4, h(4)) ) and ( (6, h(6)) ) is ( square ).
(type your answer in slope - intercept form.)

Explanation:

Step1: Calculate \( h(4) \) and \( h(6) \)

Given \( h(x)=x^{2}-2x \).
For \( x = 4 \):
\( h(4)=4^{2}-2\times4=16 - 8=8 \)
For \( x = 6 \):
\( h(6)=6^{2}-2\times6=36-12 = 24 \)

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 \( a = 4 \), \( b=6 \), \( h(4)=8 \), \( h(6)=24 \).
\( \frac{h(6)-h(4)}{6 - 4}=\frac{24 - 8}{2}=\frac{16}{2}=8 \)

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 \( m=\frac{h(6)-h(4)}{6 - 4}=8 \) (from step 2).
Using the point - slope form \( y - y_{1}=m(x - x_{1}) \), take the point \( (x_{1},y_{1})=(4,8) \)
\( y - 8=8(x - 4) \)
Expand: \( y-8=8x-32 \)
Add 8 to both sides: \( y=8x-24 \)

Answer:

(a) The average rate of change from 4 to 6 is \( 8 \).
(b) An equation of the secant line containing \( (4,h(4)) \) and \( (6,h(6)) \) is \( y = 8x-24 \).