Answer
The equation of the tangent line is $y=1/6x+5/3$.
Work Step by Step
The slope $m$ of the tangent line of the curve $f(x)$ at $A(a,b)$ is also the slope of the curve at that point, which is calculated by $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$
$$f(x)=\sqrt {x+1}\hspace{1cm}A(8,3)$$
1) Find the slope $m$ of the tangent:
$$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$
Here $a=8$ and $f(a)=b=3$
$$m=\lim_{h\to0}\frac{f(h+8)-3}{h}=\lim_{h\to0}\frac{\sqrt{h+8+1}-3}{h}=\lim_{h\to0}\frac{\sqrt{h+9}-3}{h}$$
Multiply both numerator and denominator by $\sqrt{h+9}+3$:
$$m=\lim_{h\to0}\frac{(\sqrt{h+9}-3)(\sqrt{h+9}+3)}{h(\sqrt{h+9}+3)}=\lim_{h\to0}\frac{h+9-9}{h(\sqrt{h+9}+3)}$$ $$m=\lim_{h\to0}\frac{h}{h(\sqrt{h+9}+3)}=\lim_{h\to0}\frac{1}{\sqrt{h+9}+3}$$
$$m=\frac{1}{\sqrt{0+9}+3}=\frac{1}{\sqrt9+3}=\frac{1}{6}$$
2) Find the equation of the tangent line at $A(8,3)$:
The tangent line would have this form: $$y=\frac{1}{6}x+m$$
Substitute $A(8,3)$ here to find $m$:
$$\frac{1}{6}\times8+m=3$$ $$\frac{4}{3}+m=3$$ $$m=\frac{5}{3}$$
So the equation of the tangent line is $y=1/6x+5/3$.
