Question

homework4: problem 5 (1 point) let (f = 2x^{2}+4x + 2) and find the values below 1. (f(x + h)=) 2. ((f(x + h)-f(x))=) 3. (lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}=) 4. find the equation of the line tangent to the graph of (f) at (x=-3). (y=) note: you can earn partial credit on this problem. preview my answers submit answers you have attempted this problem 0 times. you have unlimited attempts remaining. email instructor

Explanation:

Step1: Find $f(x + h)$

Substitute $x+h$ into $f(x)=2x^{2}+4x + 2$.
\[

$$\begin{align*} f(x + h)&=2(x + h)^{2}+4(x + h)+2\\ &=2(x^{2}+2xh+h^{2})+4x + 4h+2\\ &=2x^{2}+4xh+2h^{2}+4x + 4h+2 \end{align*}$$

\]

Step2: Find $f(x + h)-f(x)$

\[

$$\begin{align*} f(x + h)-f(x)&=(2x^{2}+4xh+2h^{2}+4x + 4h+2)-(2x^{2}+4x + 2)\\ &=2x^{2}+4xh+2h^{2}+4x + 4h+2 - 2x^{2}-4x - 2\\ &=4xh+2h^{2}+4h \end{align*}$$

\]

Step3: Find $\lim_{h

ightarrow0}\frac{f(x + h)-f(x)}{h}$
\[

$$\begin{align*} \lim_{h ightarrow0}\frac{f(x + h)-f(x)}{h}&=\lim_{h ightarrow0}\frac{4xh+2h^{2}+4h}{h}\\ &=\lim_{h ightarrow0}(4x + 2h+4)\\ &=4x+4 \end{align*}$$

\]

Step4: Find the equation of the tangent - line at $x=-3$

First, find the slope of the tangent - line at $x = - 3$. Substitute $x=-3$ into the derivative. The slope $m=4(-3)+4=-8$.
Next, find the value of $y$ when $x=-3$. $f(-3)=2(-3)^{2}+4(-3)+2=2\times9-12 + 2=18-12 + 2=8$.
Using the point - slope form of a line $y - y_{1}=m(x - x_{1})$ with $(x_{1},y_{1})=(-3,8)$ and $m=-8$.
\[

$$\begin{align*} y-8&=-8(x + 3)\\ y-8&=-8x-24\\ y&=-8x-16 \end{align*}$$

\]

Answer:

1. $2x^{2}+4xh+2h^{2}+4x + 4h+2$
2. $4xh+2h^{2}+4h$
3. $4x + 4$
4. $y=-8x-16$