let ( h(x)=x^{2}-6x ).\n(a) find the average rate of change from 5 to 8.\n(b) find an equation of the secant…

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

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

Answer

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 )