Tangent line at a specific point? Confused, please help

Question

What is the equation of the tangent line at $x=8$, assuming that $f(8)=3$ and $f^{'}(8)= -\frac{4}{5}$

Answer 1

Hint: A line is determined once you know a point on the line and you know the slope of the line. From the information given, what point is on the tangent line? And what is its slope?

Answer 2

The tangent line to a function $f(x)$ at a point $a$ is a linear function (i.e., a line) $\ell(x)$ which matches the function value at $a$ and slope at $a$. That is, $\ell(a)=f(a)$ and $\ell'(a)=f'(a)$. This linear function must be given by the equation $$\ell(x)=f(a)+f'(a)(x-a).$$ Why is this the right formula? Well $\ell(x)$ is linear, so its slope is just the coefficient of $x$. We want that to be $f'(a)$, so we must have $\ell(x)=f'(a)x+C$ for some constant $C$. We could also write this as $\ell(x)=f'(a)(x-a)+D$ for some other constant $D$. Now we also want $\ell(a)=f(a)$. If we plug in $a$ we get $\ell(a)=f'(a)(a-a)+D=D$, so $D$ must be $f(a)$.

Now you can use the formula for the tangent line and the information you are given to determine the answer.