Answer
$L(x)=-\frac{4}{5}x+\frac{9}{5}$
Work Step by Step
Step 1. $f(x)=\sqrt {x^2+9}=(x^2+9)^{1/2}$, $f'(x)=\frac{1}{2}(x^2+9)^{-1/2}(2x)=\frac{x}{\sqrt {x^2+9}}$,
Step 2. $f(-4)=\sqrt {(-4)^2+9}=5$, $f'(-4)=\frac{-4}{\sqrt {(-4)^2+9}}=-\frac{4}{5}$
Step 3. The linearization at $x=-4$ is $L(x)=f(-4)+f'(-4)(x+4)=5-\frac{4}{5}(x+4)=-\frac{4}{5}x+\frac{9}{5}$
