How to find the equation of the tangent to the parabola $y = x^2$ at the point (-2, 4)? This question is from George Simmons' Calc with Analytic Geometry. This is how I solved it, but I can't find the two points that satisfy this equation: 
$$
\begin{align}
\text{At Point P(-2,4):} \hspace{30pt} y &= x^2 \\
\frac{dy}{dx} &= 2x^{2-1} \\
&= 2x = \text{Slope at P.}
\end{align}
$$
Now, the equation for any straight line is also satisfied for the tangent:
$$
\begin{align}
y - y_0 &= m(x - x_0) \\
\implies y - y_0 &= 2x (x - x_0) \\
\text{For point P, } x_0 &= -2 \text{ and } y_0 = 4 \\
\implies y - 4 &= 2x(x+2)\\
\implies y - 4 &= 2x^2 + 4x\\
\implies y &= 2x^2 + 4x +4\\
\end{align}
$$
This is where the problem occurs. If I were to try to solve for $y$ using:
$$
y = ax^2+bx+c \implies y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
I'd get:
$$
\begin{align}
y &= 2x^2+4x+4 \text{ and, at x-intercept: }\\ 
x &= \frac{-4 \pm \sqrt{4^2 - (4\times2\times4)}}{2\times2} \\
x &= \frac{-4 \pm \sqrt{16 - 32}}{4} \\
x &= \frac{-4 \pm 4i}{4} \\
x &= -1 \pm i
\end{align}
$$
Is this the correct direction, or did I do something wrong? 

The others did point out your error, so I will just add the way I'd do it:
The tangent line we are looking for is in the form of $$g(x)=ax+b$$ for the function $$f(x)=x^2$$ at $x=-2$. We know that their derivate and their value most be equal at the given point, so we have that $$a=2*(-2)=-4$$ and $$(-2)^2=-4(-2)+b$$ $$4=8+b$$ $$b=-4$$ So the eqution for the tangent line is $$y=-4x-4$$
I like this method because I do not need to remember to the equation of the line through a given point.
