Question

In Problems 9–20, find the slope of the tangent line to the graph of f at the given point. Graphf and the tangent line.

$f\left(x\right)=-2x^2+x-3$at$(1,-4)$

Short answer

The slope of tangent is $-3$.

The graph is as follows,

Step-by-step solution

Step 1. Given Information

We are given a function,

$f\left(x\right)=-2x^2+x-3$

We have to find the slope of the tangent line and graph the function and tangent line.

Step 2. Finding the slope

The slope is given by,

$m_{\tan }=\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$

Putting the values, we get

role="math" localid="1647076630300" $$\begin{gathered} m=\underset{x\to 1}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-\left(1\right)} \\ =\underset{x\to 1}{\lim }\frac{-2x^2+x-3-(-4)}{x-1} \\ =\underset{x\to 1}{\lim }\frac{-2x^2+x-3+4}{x-1} \\ =\underset{x\to 1}{\lim }\frac{-2x^2+x+1}{x-1} \\ =\underset{x\to 1}{\lim }\frac{(-2x-1)(x-1)}{x-1} \end{gathered}$$

Step 3. Finding the limit

Finding the limit,

$$\begin{gathered} m=\underset{x\to 1}{\lim }(-2x-1) \\ m=-2\left(1\right)-1 \\ m=-2-1 \\ m=-3 \end{gathered}$$

Hence the slope of tangent is $-3$.

Step 4. Graphing the function

The equation of tangent is given by,

$$\begin{gathered} y-y_1=m\left(x-x_1\right) \\ y-(-4)=-3(x-1) \\ y+4=-3x+3 \\ y=3x-1 \end{gathered}$$

The graph is as follows,