SOLUTION: Find x coordinate of the point on the graph y=square root of x+8 where the tangent line is parallel to the secant line that cuts the curve at x =-8 and x=7.

Question 1180829: Find x coordinate of the point on the graph y=square root of x+8 where the tangent line is parallel to the secant line that cuts the curve at x =-8 and x=7.
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Evaluate the function at x=-8 and x=7: f(-8)=0; f(7)=sqrt(15).

Find the equation of the secant: (-8,0) to (7,sqrt(15))

slope = sqrt(15)/(7-(-8)) = sqrt(15)/15

Graph the function and the secant:

Find the derivative of the function:

Find the x value where the slope of the tangent is the same as the slope of the secant.

Graph the line with slope sqrt(15)/15 touching y=sqrt(x+8) at x=-17/4 and observe that it is parallel to the secant:

Graph the function and the secant:

ANSWER: x = -17/4.

At x=-17/4, the tangent to y=sqrt(x+8) is parallel to the secant through x=-8 and x=7.

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