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}}=(2^2-(-1)^2)/(2-(-1))=\frac33=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=\frac12`.