Answer
$L(x)=\frac{-4}{5}x+\frac{9}{5}$
Work Step by Step
Given: $f(x)=\sqrt{(x^2+9)}$,
L(x)=f(a)+f'(a)(x-a)
$f'(x)=\frac{1}{2\sqrt{(x^2+9)}}\frac{d{(x^2+9)}}{dx}$
$f'(x)=\frac{x}{\sqrt{(x^2+9)}}$
a=-4
$f(-4)=\sqrt{((-4)^2+9)}=5$
$f'(-4)=\frac{-4}{\sqrt{((-4)^2+9)}}=\frac{-4}{5}$
then $L(x)=5+\frac{-4}{5}(x+4)$
$L(x)=5+\frac{-4}{5}x-\frac{16}{5}=\frac{-4}{5}x+\frac{9}{5}$
$L(x)=\frac{-4}{5}x+\frac{9}{5}$
Thus: $L(x)=\frac{-4}{5}x+\frac{9}{5}$
