Answer
$m=24$
Work Step by Step
$f(x)=2x^{3},$ at $(2,16)$
The slope of the tangent line at the point $P(a,f(a))$ is given by $m=\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h}$
In this case, $a=2$
Find $f(a+h)$ by substituting $x$ by $2+h$ in $f(x)$ and simplifying:
$f(2+h)=2(2+h)^{3}=2(2^{3}+3(2)^{2}h+3(2)h^{2}+h^{3})=...$
$...=2(8+12h+6h^{2}+h^{3})=2h^{3}+12h^{2}+24h+16$
Since $f(a)$ is the $y$-coordinate of the point given, $f(a)=16$
Substitute the known values into the formula that gives the slope of the tangent line and evaluate:
$m=\lim_{h\to0}\dfrac{2h^{3}+12h^{2}+24h+16-16}{h}=...$
$...=\lim_{h\to0}\dfrac{2h^{3}+12h^{2}+24h}{h}=\lim_{h\to0}\dfrac{h(2h^{2}+12h+24)}{h}=...$
$...=\lim_{h\to0}(2h^{2}+12h+24)=2(0)^{2}+12(0)+24=24$
