let ( h(x)=x^{2}-7x ).\n(a) find the average rate of change from 5 to 7.\n(b) find an equation of the secant…

let ( h(x)=x^{2}-7x ).\n(a) find the average rate of change from 5 to 7.\n(b) find an equation of the secant line containing ( (5, h(5)) ) and ( (7, h(7)) ).\n(a) the average rate of change from 5 to 7 is ( square ). (simplify your answer.)\n(b) an equation of the secant line containing ( (5, h(5)) ) and ( (7, h(7)) ) is ( square ).\n(type your answer in slope - intercept form.)

let ( h(x)=x^{2}-7x ).\n(a) find the average rate of change from 5 to 7.\n(b) find an equation of the secant line containing ( (5, h(5)) ) and ( (7, h(7)) ).\n(a) the average rate of change from 5 to 7 is ( square ). (simplify your answer.)\n(b) an equation of the secant line containing ( (5, h(5)) ) and ( (7, h(7)) ) is ( square ).\n(type your answer in slope - intercept form.)

Answer

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 )