Find the equation of the tangent line(s) to the curve $f(x) = \sqrt{x + 5}$, that passes through the point $(-8, 1$). [$-8$ can't get subbed in].

Question

Going through the process of finding the derivative, which is $\dfrac{1}{2\sqrt{x+5}}$ ,and substituting x gives an undefined slope, as you end up taking the square root of a negative number

I know that the point exists somewhere across the tangent line, but I don't know how to translate this into finding an answer.

Answer 1

The tangent line to the point $(a,f(a))$ is given by $$y(x) = f(a)+ f'(a) (x-a) = \sqrt{a+5} + \frac{x-a}{2 \sqrt{a+5}}$$

You want to find a point $a$ whose tangent line crosses $(-8,1)$, so you need to solve $$ 1 = \sqrt{a+5} + \frac{-8-a}{2 \sqrt{a+5}} $$ whose acceptable solution is $a=4$

There is no square root problem since the tangent line is defined over all reals, even though the function you started with wasn't

Answer 2

The derivative will give the slope of the tangent line for any point on the graph. The derivative of $\sqrt{x + 5}$ is $\dfrac{1}{2\sqrt{x+5}}$. This slope can also be written using the original equation and point as: $\frac{\sqrt{x+5}-1}{x+8}$. Setting these equal you can find that $x = 4$, which means the slope is $\frac{1}{6}$, and the equation is $f(x) = y = \frac{1}{6}x+\frac{5}{3}$. Might have made a mistake somewhere but this is the way you solve it.