Answer
$y=\dfrac{1}{3}x+\dfrac{5}{3}$
The graph of the function and the tangent line is shown below:
Work Step by Step
$y=\sqrt{1+2x},$ at $(4,3)$
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, $a=4$
Find $f(a+h)$ by substituting $x$ by $4+h$ in $f(x)$ and simplifying:
$f(4+h)=\sqrt{1+2(4+h)}=\sqrt{1+8+2h}=\sqrt{9+2h}$
Since $f(a)$ is the $y$-coordinate of the point given, $f(a)=3$
Substitute the known values into the formula that gives the slope of the tangent line and evaluate:
$m=\lim_{h\to0}\dfrac{\sqrt{9+2h}-3}{h}\cdot\dfrac{\sqrt{9+2h}+3}{\sqrt{9+2h}+3}=...$
$...=\lim_{h\to0}\dfrac{(\sqrt{9+2h})^{2}-3^{2}}{h(\sqrt{9+2h}+3)}=\lim_{h\to0}\dfrac{9+2h-9}{h(\sqrt{9+2h}+3)}=...$
$...=\lim_{h\to0}\dfrac{2h}{h(\sqrt{9+2h}+3)}=\lim_{h\to0}\dfrac{2}{\sqrt{9+2h}+3}=...$
$...=\dfrac{2}{\sqrt{9+2(0)}+3}=\dfrac{2}{3+3}=\dfrac{2}{6}=\dfrac{1}{3}$
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:
$y-3=\dfrac{1}{3}(x-4)$
$y-3=\dfrac{1}{3}x-\dfrac{4}{3}$
$y=\dfrac{1}{3}x-\dfrac{4}{3}+3$
$y=\dfrac{1}{3}x+\dfrac{5}{3}$
