Answer
$y=\dfrac{1}{4}x+\dfrac{7}{4}$
The graph of the function and the tangent line is shown below:
Work Step by Step
$y=\sqrt{x+3},$ at $(1,2)$
The slope of the tangent line at the point $P(a,f(a))$ is given by $m=\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h}$
In this case, $a=1$
Find $f(a+h)$ by substituting $x$ by $1+h$ in $f(x)$ and simplifying:
$f(1+h)=\sqrt{1+h+3}=\sqrt{4+h}$
Since $f(a)$ is the $y$-coordinate of the point given, $f(a)=2$
Substitute the known values into the formula that gives the slope of the tangent line and evaluate:
$m=\lim_{h\to0}\dfrac{\sqrt{4+h}-2}{h}\cdot\dfrac{\sqrt{4+h}+2}{\sqrt{4+h}+2}=...$
$...=\lim_{h\to0}\dfrac{(\sqrt{4+h})^{2}-2^{2}}{h(\sqrt{4+h}+2)}=\lim_{h\to0}\dfrac{4+h-4}{h(\sqrt{4+h}+2)}=...$
$...=\lim_{h\to0}\dfrac{h}{h(\sqrt{4+h}+2)}=\lim_{h\to0}\dfrac{1}{\sqrt{4+h}+2}=...$
$...=\dfrac{1}{\sqrt{4+0}+2}=\dfrac{1}{2+2}=\dfrac{1}{4}$
The point-slope form of the equation of a line is $y-y_{0}=m(x-x_{0})$. Both the slope of the line and a point through which it passes are known. Substitute them into the formula to obtain the equation of the tangent line:
$y-2=\dfrac{1}{4}(x-1)$
$y-2=\dfrac{1}{4}x-\dfrac{1}{4}$
$y=\dfrac{1}{4}x-\dfrac{1}{4}+2$
$y=\dfrac{1}{4}x+\dfrac{7}{4}$