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Find the x-coordinate of the point on the graph of y=x^2 where the tangent line is parallel to the secant line that cuts the curve at x=-1 and x=2

Please how do you solve this?

Verified Answer :

LammettHash

First find the secant line. The slope of the secant line through \((-1,1)\) (when \(x=-1\)) and \((2,4)\) (when \(x=2\)) is the average rate of change of \(y=x^2\) over the interval \([-1,2]\):

\(\text{slope}_{\text{secant}}=\dfrac{2^2-(-1)^2}{2-(-1)}=\dfrac33=1\)

The tangent line to \(y=x^2\) will have a slope determined by the derivative:

\(y=x^2\implies y'=2x\)

Both the secant and tangent will have the same slope when \(2x=1\), or when \(x=\dfrac12\).