given the function $f(x)=-2x^{2}$
find the difference quotient $\frac{f(x + h)-f(x)}{h}$
find $f(x)$ by determining $lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$
use the previous answer to find the values indicated below. the values should match the slopes of the tangent lines you will draw on the graph at the bottom.
find $f(-2)$
find $f(0)$
find $f(1)$
graph the function $f(x)=-2x^{2}$ and draw the tangent lines to the graph at points whose x - coordinates are - 2, 0, and 1.

Question

Accepted Answer

# Explanation:
## Step1: Find the difference quotient
First, find $f(x + h)$:
$$
\begin{align*}
f(x+h)&=-2(x + h)^{2}\
&=-2(x^{2}+2xh+h^{2})\
&=-2x^{2}-4xh - 2h^{2}
\end{align*}
$$
Then, calculate $\frac{f(x + h)-f(x)}{h}$:
$$
\begin{align*}
\frac{f(x + h)-f(x)}{h}&=\frac{-2x^{2}-4xh - 2h^{2}-(-2x^{2})}{h}\
&=\frac{-2x^{2}-4xh - 2h^{2}+2x^{2}}{h}\
&=\frac{-4xh - 2h^{2}}{h}\
&=-4x-2h
\end{align*}
$$
## Step2: Find the derivative $f^{\prime}(x)$
$$
\begin{align*}
f^{\prime}(x)&=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}\
&=\lim_{h
ightarrow0}(-4x - 2h)\
&=-4x
\end{align*}
$$
## Step3: Find $f^{\prime}(-2)$
Substitute $x=-2$ into $f^{\prime}(x)$:
$$f^{\prime}(-2)=-4\times(-2)=8$$
## Step4: Find $f^{\prime}(0)$
Substitute $x = 0$ into $f^{\prime}(x)$:
$$f^{\prime}(0)=-4\times0 = 0$$
## Step5: Find $f^{\prime}(1)$
Substitute $x = 1$ into $f^{\prime}(x)$:
$$f^{\prime}(1)=-4\times1=-4$$
# Answer:
Difference quotient: $-4x - 2h$
$f^{\prime}(x)$: $-4x$
$f^{\prime}(-2)$: $8$
$f^{\prime}(0)$: $0$
$f^{\prime}(1)$: $-4$

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QUESTION IMAGE

Question

Explanation:

Step1: Find the difference quotient

Step2: Find the derivative \(f^{\prime}(x)\)

Step3: Find \(f^{\prime}(-2)\)

Step4: Find \(f^{\prime}(0)\)

Step5: Find \(f^{\prime}(1)\)

Answer: