How do you find a tangent line parallel to secant line?

Question

Ann Cole · Accepted Answer

The Mean Value Theorem may be used to locate a tangent line that is parallel to a secant line.
According to the Mean Value Theorem, if a function is continuous and differentiable, then
$f^{'} (x) = \frac{f (b) - f (a)}{b - a}$
This formula requires a function f. (x). I'll apply $f (x) = - x^{3}$ as an example.
Additionally, I'll use a=-2 and b=2 for the secant line's interval. The line that connects (-2, 8) and is shown here (2,-8).
So, we know that the slope of this line will be $\frac{- 8 - 8}{2 - (- 2)} = - 4$ .
We will take the derivative of the function, f'(x), set it to -4, then solve for x in order to get the tangent lines parallel to this secant line.
$- 3 x^{2} = - 4$
Solving this for x gives us: $x = \pm \sqrt{\frac{4}{3}}$ .
So, the lines tangent to $y = - x^{3}$ at $x = \sqrt{\frac{4}{3}}$ and $x = - \sqrt{\frac{4}{3}}$ parallel to the secant line via x=2 and x=-2 is required.

How do you find a tangent line parallel to secant

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