Let $f(x) =14-x^2$
The slope of the tangent line to the graph of $f(x)$ at the point $(-3,5)$ is .
The equation of the tangent line to the graph of $f(x)$ at $(-3,5)$ is $y=mx+b$ for
$m=$
and
$b=$.
Hint: the slope at $x=-3$ is given by
$\displaystyle{ m = \lim_{h\to 0}\frac{f(-3+h)-f(-3)}{h}}$