$$-2x^2-15x-25\over x^2-x-5$$
I'm not sure how you would find the Intercepts or the Asymptote of this function.
I've tried factoring the equation but it leaves me with
$$-1(2x+5)(x+5)\over x^2-x-5$$
But I'm confused, how would I find the key features from there
On
I think key features are zeros, poles and asymptotes?
For the expression to be zero the numerator must be zero while at that point while the denominator must be nonzero. As you already noticed, the numerator is zero fo $x=-5/2$ and $x=-5$ and the denominator is nonzero at both those points, so these are the zeros.
Then the poles are the opposite: The points where the denominator is zero but the numerator isn't. As they have no common zeros, the poles are just the zeros of the denominator.
In case you had the same linear factors in the denominator and in the numerator it would have been just a removable singularity. As the poles are basically the vertical asymptotes you now neet to find the asymptote for $x \to \pm \infty$ This can easily be done with polynomial division:
$$\frac{-2x^2-15x-25}{x^2-x-5} = -2 + \frac{-17x-35}{x^2-x-5}$$
So for $x \to \pm \infty$ the $x^2$ in the denominator will dominate the $x$ in the numerator (each the highest order terms) so we know that the this fraction will converge to zero for $x\to \pm \infty$. What will be left is $y= -2$ which is the asymptote we are looking for.
at first you have to search for the zero points of $x^2-x-5$ and then you can write $-2+{\frac {-17\,x-35}{{x}^{2}-x-5}}=\frac{-2x^2-15x-25}{x^2-x-5}$