#### Answer

$$L(x)\approx -8x +48$$

#### Work Step by Step

Given
$$ g(x)=\frac{x^{2}}{x-3}, \quad a=4$$
Since $g(4)=16$ and
\begin{align*}
g'(x)&=\frac{(x-3) 2 x-x^{2}(1 )}{(x-3)^{2}}\\ &=\frac{2 x^{2}-6 x-x^{2}}{(x-3)^{2}}\\ g'(4)&=-8 \end{align*}
Then the linearization at $a=4$ is given by
\begin{align*}
L(x)&\approx g^{\prime}(a)(x-a)+f(a)\\ &=-8\left(x-4\right)+16\\ &=-8x +48 \end{align*}