Answer
$y=-x-1$
The graph of the function and the tangent line is shown below:
Work Step by Step
$y=x+x^{2},$ at $(-1,0)$
The slope of the tangent line at the point $P(a,f(a))$ is given by $m=\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h}$
In this case, $a=-1$
Find $f(a+h)$ by substituting $x$ by $-1+h$ in $f(x)$ and simplifying:
$f(-1+h)=-1+h+(-1+h)^{2}=-1+h+1-2h+h^{2}=...$
$...=h^{2}-h$
Since $f(a)$ is the $y$-coordinate of the point given, $f(a)=0$
Substitute the known values into the formula that gives the slope of the tangent line and evaluate:
$m=\lim_{h\to0}\dfrac{h^{2}-h}{h}=\lim_{h\to0}\dfrac{h(h-1)}{h}=\lim_{h\to0}(h-1)=...$
$...=0-1=-1$
The point-slope form of the equation of a line is $y-y_{0}=m(x-x_{0})$. Both the slope of the line and a point through which it passes are given. Substitute them into the formula to obtain the equation of the tangent line:
$y-0=-1(x+1)$
$y=-x-1$