Answer
a) $-16$
b) $ y=-16x-16$
Work Step by Step
a) Now, $ m_{\tan}=\lim_\limits{ h\to 0} \dfrac{f(-2+h)-f(-2)}{h}$
or, $=\lim_\limits{ h\to 0} \dfrac{4(4-4h+h^2)-16}{h}$
or, $=\lim_\limits{ h\to 0} \dfrac{-16h+4^2}{h}$
At $(-2,16)$; $ m_{\tan}=-16$
b) The equation of a line is: $ y=mx+c $
Now, $ y-16=-16(x+2) \implies y=-16x-16$
