Chapter 14.4, Problem 11AYU

To determine

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

Answer to Problem 11AYU

$-2$

Explanation of Solution

Given:

$f\left(x\right) = x^2 + 2$

Point: $\left(-1, 3\right)$

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, $y = f\left(x\right) = x^2 + 2$

Slope of the tangent is given by;

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

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

$\frac{dy}{dx} = 2x + 0$

$\frac{dy}{dx} = 2x$

At the point $\left(-1, 3\right)$; $\frac{dy}{dx} = 2\left( - 1\right) = -2$

Hence the slope of the tangent to the curve $y = f\left(x\right) = x^2 + 2$ at $\left(-1, 3\right)$ is $-2$.

The graph is as follows: