Find slope of the tangent of a curve passing through fixed point I need to find the slope of the tangent of $f(x)=\sqrt{2x-10}$ that pass through point $(0,0)$.

The derivative of $f(x)$ is $1/(\sqrt{2x-10})$.
Observe that the given curve is a branch of a parabola. Since the tangent line passes through $(0,0)$. Therefore it is of the form $y=mx$. Now find the intersection of this with the given curve by solving
\begin{align*}
mx &amp; = \sqrt{2x-10}\\
m^2x^2 &amp; = 2x-10\\
m^2x^2-2x+10 &amp; = 0.
\end{align*}
For this to be a tangent the line should intersect the curve at coincident points, meaning the roots of the above quadratic must be equal. Thus
$$4-40m^2 =0 \implies m=\pm \frac{1}{\sqrt{10}}.$$
Keeping in mind that this is the upper branch of the parabola. $m=\frac{1}{\sqrt{10}}$.
The tangent is given by $$mx+ b,$$ with $m$ being the derivative of $f$ in $x=a$ ($x$-value to be evaluated).