Find a possible function $h\left(x\right)$ and a number $c$ such that the following limit represents the slope of the curve $y=h\left(x\right)$ at $(c,h(c\left)\right)$. Then, evaluate the limit:
$\underset{x\to -1}{\lim }\frac{4x^2-2x-6}{x+1}$

$h\left(x\right)=4x^2-2x$

$c=-1$

${\displaystyle\lim_{x\to-1}}\frac{4x^2-2x-6}{x+1}$ $=$ $10$

$h\left(x\right)=4x^2-2x$

$c=-1$

${\displaystyle\lim_{x\to-1}}\frac{4x^2-2x-6}{x+1}$ $=$ $-10$

$h\left(x\right)=4x^2+2x$

$c=-1$

${\displaystyle\lim_{x\to-1}}\frac{4x^2-2x-6}{x+1}$ $=$ $10$

$h\left(x\right)=4x^2+2x$

$c=1$

${\displaystyle\lim_{x\to-1}}\frac{4x^2-2x-6}{x+1}$ $=$ $-10$