Question

homework4: problem 8 (1 point) let f = 2x^2 + 4 and find the values below 1. f(x + h)= 2. (f(x + h)-f(x))= 3. lim_{h→0} (f(x + h)-f(x))/h = 4. find the equation of the line tangent to the graph of f at x = 5. 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 page generated october 5, 2025, 11:21:28 pm cdt

Explanation:

Step1: Find $f(x + h)$

Substitute $x+h$ into $f(x)$:
\[

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

\]

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

\[

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

\]

Step3: Find $\lim_{h

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

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

\]

Step4: Find the equation of the tangent - line at $x = 5$

First, find the slope of the tangent - line at $x = 5$. Substitute $x = 5$ into the derivative. The derivative $f^\prime(x)=4x$, so $m=f^\prime(5)=20$.
Next, find the $y$ - value when $x = 5$. $f(5)=2\times5^{2}+4=2\times25 + 4=50 + 4=54$.
Use the point - slope form of a line $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(5,54)$ and $m = 20$.
\[

$$\begin{align*} y-54&=20(x - 5)\\ y-54&=20x-100\\ y&=20x - 46 \end{align*}$$

\]

Answer:

1. $2x^{2}+4xh + 2h^{2}+4$
2. $4xh+2h^{2}$
3. $4x$
4. $y = 20x-46$