Given the following problem:
Let $f$ be the real-valued function defined by $f(x) = \sqrt{1+6x}$.
Determine the slope of the line tangent to the graph of $f$ at $x=4$.
Determine the y-intercept of the line tangent to the graph of $f$ at $x=4$.
For the life of my I can't work out why the solution to the $y$ intercept is given as $\frac{13}{5}$.
I've correctly solved $f'$ using the definition of a derivative formula (I got $\frac{3}{\sqrt{1+6x}}$) and do get the correct answer for the slope of the line tangent to $x=4$ (which is $\frac{3}{5}$) but the $y$ intercept for that line is definitely $3$ and not $\frac{13}{5}$. Where am I going wrong? This is the solution that is given by the authors:
Say the tangent line is $y=mx+b$ with $m=\frac35$ and a point on it is $(x,y)=(4,5).$
Can you solve for $b$?