Find the equation of all tangents to the curve $y=-3x^2-5$ from the point $(-4,-5)$

Question

I'm not sure where to start with this question other then finding the derivative and finding the slope. The question asking for all tangents is confusing me. Any help is appreciated. For reference, I am a Grade $12$ student.

Answer 1

hint: Let $y + 5 = m(x+4) $ be the tangent line equation. Thus $y = mx+4m-5$. Let it equal to $y = 3x^2-5$, and solve $\triangle = b^2-4ac = 0$ to get all the $m$'s . Does this help?

Answer 2

Let the equation of the tangent line be $y=m(x+4)-5.$ At the point of tangency, we have $y=-3x^2-5$, so at this point, $-3x^2=m(x+4).$ We also know that $m=-6x$ so again at the point of tangency, $-3x^2=-6x(x+4).$ Now this has two solutions $x=0\space (m=0)$ and $x=-8\space (m=48)$ which give you the equations of two lines.