Answer
The x-coordinate of the point on the graph of $y=x^2$ where the tangent line is parallel to the secant line is $x=\frac{1}{2}$.
Work Step by Step
We begin by calculating the slope of the secant line. Note that the secant line crosses the curve at $(-1,1)$ and $(2,4)$. Therefore, the slope of the secant line is $\frac{\Delta y}{\Delta x} = \frac{4-1}{2-(-1)} = \frac{3}{3}=1$.
To find the x-coordinate of the point on the graph of $y=x^2$ where the tangent line is parallel to the secant line, we find at the coordinate in which the instantaneous slope or derivative equals the secant line's slope. We apply the power rule to find the derivative of $y$.
$y' = 2x$
We equate this to the slope of the secant line and solve for $x$:
$y' = 2x = 1$
$x=\frac{1}{2}$.
Therefore, the x-coordinate of the point on the graph of $y=x^2$ where the tangent line is parallel to the secant line is $\frac{1}{2}$.
