let ( h(x)=x^{2}-2x ).\n(a) find the average rate of change from 4 to 6.\n(b) find an equation of the secant…

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

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

Answer

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 ).