Answer
$f'(a)=2$
Equation of the tangent line: $y=2x+5$
Work Step by Step
Recall that $f'(a)=\lim\limits_{h \to 0}\frac{f(a+h)-f(a)}{h}$
Given that $a=-1$. Therefore,
$f'(-1)=\lim\limits_{h \to 0}\frac{f(-1+h)-f(-1)}{h}$
$=\lim\limits_{h \to 0}\frac{[4-(-1+h)^{2}]-[4-(-1)^{2}]}{h}=\lim\limits_{h \to 0}\frac{[4-(1+h^{2}-2h)]-3}{h}$
$=\lim\limits_{h \to 0}\frac{4-1-h^{2}+2h-3}{h}=\lim\limits_{h \to 0}\frac{-h^{2}+2h}{h}=\lim\limits_{h \to 0}(-h+2)=2$
Equation of a tangent line is in the form
$y-f(a)=f'(a)(x-a)$
Knowing that $f(a)=f(-1)=3,$ $f'(a)=f'(-1)=2$ and $a=-1$,
we obtain the equation of the tangent line through $a$ as below.
$y-3=2(x-(-1))$
Or $y=2x+5$
