Question

1. \( f(x)=\sqrt{x+3} \) a. Find the slope of the tangent line at \( a=1 \) b. Find the equation of the tangent line at \( a=1 \) \[ \begin{array}{l} \text { Use } m_{\text {tan }}=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h} \\ m_{\text {fan }}=\lim _{h \rightarrow 0} \frac{\sqrt{4+h}-2}{h} \cdot \frac{\sqrt{4+h}+2}{\sqrt{4+h}+2}=\lim _{n \rightarrow 0} \frac{4+h-4}{h(\sqrt{4+n}+2)}=\lim _{h \rightarrow 0} \frac{1}{\sqrt{4+h}+2}=\frac{1}{4} \\ y-2=\frac{1}{4}(x-1) \\ y=\frac{1}{4} x-\frac{1}{4}+\frac{8}{4} \end{array} \]

          1. \( f(x)=\sqrt{x+3} \)
a. Find the slope of the tangent line at \( a=1 \)
b. Find the equation of the tangent line at \( a=1 \)
\[
\begin{array}{l}
\text { Use } m_{\text {tan }}=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h} \\
m_{\text {fan }}=\lim _{h \rightarrow 0} \frac{\sqrt{4+h}-2}{h} \cdot \frac{\sqrt{4+h}+2}{\sqrt{4+h}+2}=\lim _{n \rightarrow 0} \frac{4+h-4}{h(\sqrt{4+n}+2)}=\lim _{h \rightarrow 0} \frac{1}{\sqrt{4+h}+2}=\frac{1}{4} \\
y-2=\frac{1}{4}(x-1) \\
y=\frac{1}{4} x-\frac{1}{4}+\frac{8}{4}
\end{array}
\]

1. f(x)=√(x+3)
a. Find the slope of the tangent line at a=1
b. Find the equation of the tangent line at a=1

Use mtan =limh → 0(f(a+h)-f(a))/(h)

mfan =limh → 0(√(4+h)-2)/(h)·(√(4+h)+2)/(√(4+h)+2)=limn → 0(4+h-4)/(h(√(4+n)+2))=limh → 0(1)/(√(4+h)+2)=(1)/(4)

y-2=(1)/(4)(x-1)

y=(1)/(4) x-(1)/(4)+(8)/(4)

Added by Phillip A.

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