Equation of the tangent to a graph where the point is not on the graph. If $f(x)=(x+1)^{3/2}$, provided $x\geq -1$, I am asked to find the equation of all tangent lines to $f(x)$ at the point $(\frac{4}{3},3)$.
Simple enough. I first took the derivative which is:
$$f'(x)= \frac {3\sqrt{x+1}}{2}$$
Since $(\frac{4}{3},3)$ is not on the graph of $f(x)$, I need to find the point of tangency.
So I assumed the point of tangency is at some point $(k,(k+1)^{\frac{3}{2}})$ It also has to pass the point $(\frac{4}{3},3)$ so thus:
$y-y_{1} = m(x-x_{1})$
$3-(k+1)^{\frac{3}{2}} = \frac{3}{2}\sqrt{k+1}(\frac{4}{3}-k)$
$3-\sqrt{(k+1)^2(k+1)} = 2\sqrt{k+1}-\frac{3}{2}k\sqrt{k+1}$
$3-(k+1)\sqrt{k+1} = 2\sqrt{k+1}-\frac{3}{2}k\sqrt{k+1}$
$3-k\sqrt{k+1}-\sqrt{k+1}=2\sqrt{k+1}-\frac{3}{2}k\sqrt{k+1}$
$3+\frac{1}{2}k\sqrt{k+1}-3\sqrt{k+1}=0$
Here is where I am stuck. How do I solve for $k$ (And hence $x$)?
I tried letting $\sqrt{k+1}$ equal $a$ and then try and solve the corresponding cubic but that's not getting me anywhere... Any help on this? Thanks!

HINT
Use the fact that your line passes through $(k,f(k))$ and $(4/3,3)$, so the slope must also be
$$
m = \frac{\Delta y}{\Delta x} = \frac{3 - f(k)}{4/3 - k}
$$
and an alternate equation would be
$$
y - 3 = m(x - 4/3)
$$
but the equations have to match exactly...
