Question

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?

Answer 1

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 $x=\dfrac12$.