Chapter 14.4, Problem 14AYU

To determine

To find: The slope of the tangent to the function $y = f\left(x\right)$ at the indicated point.

Answer to Problem 14AYU

16

Explanation of Solution

Given:

$f\left(x\right) = -4x^2$

Point: $\left(-2, -16\right)$

Formula used:

The slope of the tangent to the curve $y = f\left(x\right)$ is the derivative of $f\left(x\right)$ w.r.to $x$ ie $\frac{dy}{dx}$.

Calculation:

Here, $y = f\left(x\right) = -4x^2$

Slope of the tangent is given by;

$\frac{dy}{dx} = \frac{d}{dx}\left(-4x^2\right)$

$\frac{dy}{dx} = \left(-4\right) . \frac{d}{dx}\left(x^2\right)$

$\frac{dy}{dx} = \left(-4\right) . \left(2x\right)$

$\frac{dy}{dx} = -8x$

At the point $\left(-2, -16\right)$; $\frac{dy}{dx} = -\mathrm{8 . }\left(-2\right) = 16$

Hence the slope of the tangent to the curve $y = f\left(x\right) = -4x^2$ at $\left(-2, -16\right)$ is 16.

The graph is as follows: