<s>[INST] How to find the equation of the tangent line to $y=x^2+2x-4$ at $x=2$? 
I'm given a curve $$y=x^2+2x-4$$
How do I find the tangent line to this curve at $x = 2$?

 [/INST] Here's an algebraic approach that avoids the explicit use of derivatives. 
We are given a quadratic function $f(x) = x^2 + 2x -4$, and we want to find the equation of the tangent to the parabola $y = f(x)$ at the point $(2, 4)$. (Note that $f(2) = 2^2 + 2 \cdot 2 - 4 = 4$.) Assume that it is given by the equation
$$
y = m(x-2) + 4, \tag{$\ast$}
$$
where $m$ is its slope.
Let's consider the intersection of the parabola with the tangent; this is given by the system of equations
$$
\begin{cases}
y &=& x^2 + 2x - 4,
\\ y &=& m(x-2)+4. 
\end{cases}
$$
In other words, to find the intersection, we should solve the quadratic  equation
$ x^2 + 2x - 4 = m(x-2)+4$, or
$$ 
x^2 + (2-m)x+(2m-8) = 0. \tag{$\ast\ast$}
$$
using the quadratic formula like so
$$
\frac{-(2-m)\pm\sqrt{(2-m)^2-4.1.(2m-8)}}{2.1}
$$
Pictorially it is clear that the tangent meets the parabola meets in exactly one point, so we want $(\ast \ast)$ to have a unique solution. This implies that the discriminant of $(\ast \ast)$ vanishes: 
$$
\begin{array}{crcl}
&(2-m)^2 - 4 \cdot (2m-8) &=& 0 
\\ \implies \qquad & m^2 - 12m + 36 &=& 0.
\end{array}
$$
Conveniently (although this is not a numerical coincidence), this equation has a unique solution $m=6$: this is the solution we are after. Plugging $m=6$ in $(\ast)$, we get the equation of the tangent at $(2, 4)$ to be $y = 6x-8$.
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