Question

a. use the definition $m_{tan}=lim_{h \to 0}\frac{f(a + h)-f(a)}{h}$ to find the slope of the line tangent to the graph of $f$ at $p$. b. determine an equation of the tangent line at $p$. $f(x)=x^{2}+3$, $p(-5,28)$ a. when using the definition $m_{tan}=lim_{h \to 0}\frac{f(a + h)-f(a)}{h}$, which is an expression for the limit to use to find the slope of the tangent line at $p$? a. $lim_{h \to 0}\frac{f(-5)-f(-5 + h)}{h}$ b. $lim_{h \to 0}\frac{f(-5 + h)+f(-5)}{h - a}$ c. $lim_{h \to 0}\frac{f(-5 + h)-f(-5)}{h}$ d. $lim_{h \to 0}\frac{f(-5 + h)+f(-5)}{h}$

Explanation:

Step1: Recall the slope - definition formula

The formula for the slope of the tangent line using the definition is $m_{tan}=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}$. Here, $a=-5$ and $f(x)=x^{2}+3$.

Step2: Substitute into the formula

We substitute $a = - 5$ into the formula. So we need to find $\lim_{h
ightarrow0}\frac{f(-5 + h)-f(-5)}{h}$.

for part b:

Step1: First, find $f(-5)$ and $f(-5 + h)$

Given $f(x)=x^{2}+3$, then $f(-5)=(-5)^{2}+3=25 + 3=28$. And $f(-5 + h)=(-5 + h)^{2}+3=25-10h+h^{2}+3=h^{2}-10h + 28$.

Step2: Calculate the slope

$m_{tan}=\lim_{h
ightarrow0}\frac{f(-5 + h)-f(-5)}{h}=\lim_{h
ightarrow0}\frac{(h^{2}-10h + 28)-28}{h}=\lim_{h
ightarrow0}\frac{h^{2}-10h}{h}=\lim_{h
ightarrow0}(h - 10)=-10$.

Step3: Use the point - slope form for the tangent line equation

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(-5,28)$ and $m=-10$. So $y - 28=-10(x + 5)$. Expand to get $y-28=-10x-50$, and then $y=-10x - 22$.

Answer:

A. $\lim_{h
ightarrow0}\frac{f(-5 + h)-f(-5)}{h}$