Answer
$$
h(x)=\left\{\begin{array}{ll}{\frac{3 x^{2}+2 x-8}{x+2}} & {\text { if } x \neq-2} \\ {3 x+k} & {\text { if } x=-2}\end{array}\right.
$$
The value of the constant $k$ that makes the function continuous is .
$k=-4 $
Work Step by Step
$$
h(x)=\left\{\begin{array}{ll}{\frac{3 x^{2}+2 x-8}{x+2}} & {\text { if } x \neq-2} \\ {3 x+k} & {\text { if } x=-2}\end{array}\right.
$$
This rational function is discontinuous wherever the denominator is zero.
There is a discontinuity when $2=-2$ .
We investigate at $x=-2$ first.
$$
\begin{aligned}
\lim _{x \rightarrow -2} h(x) &=\lim _{x \rightarrow -2}\frac{3 x^{2}+2 x-8}{x+2} \\
&=\lim _{x \rightarrow -2}\frac{(3 x-4)(x+2)}{(x+2)} \\
&=\lim _{x \rightarrow -2}(3 x-4)\\
&=3 (-2)-4 \\
&=-10
\end{aligned}
$$
Furthermore,
$$
h(-2)= 3(-2)+k=-6+k
$$
since, $h$ is continuous, then we have
$$
h(-2)=\lim _{x \rightarrow -2} h(x)
$$
so we obtain
$$
\begin{aligned} -6+k &=-10\\ k &=-10+6 \\ -4 &=k \end{aligned}
$$
Thus the value of the constant $k$ that makes the function continuous is .
$k=-4$
