Chapter 14.4, Problem 10AYU

To determine

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

Answer to Problem 10AYU

Solution:

$-2$

Explanation of Solution

Given:

$f\left(x\right)=-2x+1$

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, $y=f\left(x\right)=-2x+1$.

Slope of the tangent is given by;

$\frac{dy}{dx}=\frac{d}{dx}\left(-2x+1\right)$

$\frac{dy}{dx}=\frac{d}{dx}\left(-2x\right)+\frac{d}{dx}\left(1\right)$

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

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

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

The graph is as follows: