## Calculus (3rd Edition)

$f'(a)=2$ Equation of the tangent line: $y=2x+5$
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 $y=2x+5$