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 $x = 2$?


Another method that can be used which is traditionally taught before derivatives is the limit (they're very similar but this is more intuitive):
Let $f(x) = x^2 + 2x - 4$ and $f'(x)$ be the tangent line at any point $a$ on the curve $f(x)$
Difference quotient:
$$f'(x) =\lim_{h \rightarrow 0}\frac{f(a + h) - f(a)}{h}$$
What this essentially means is that we we take the slope between two points $a$ and $a+h$ as $h$ gets very small or as $h\rightarrow0$ the point $a+h$ gets closer to $a$ until there mathematically the slope is tangent to the point $a$.
$$f'(x) =\lim_{h \rightarrow 0} \frac{[(a+h)^2+ 2(a+h) - 4] - [(a)^2 - 2(a) - 4]}{h} $$
$$=\lim_{h \rightarrow 0} \frac{[(h^2 + 2ah + a^2) + (2a + 2h) - 4] - [a^2 - 2a - 4]}{h}$$
$$=\lim_{h \rightarrow 0} \frac{h^2 + 2ah + 2h}{h}$$
$$=\lim_{h \rightarrow 0} h + 2a + 2$$
$$= 2a + 2$$
There's your formula for the slope at any value $a$ along the curve $f(x)$
Sorry, I misread your post now I see that you want the equation!
We'll right the formula out in the form $y = mx + b$
Well we can begin with finding the slope of the line at $a = 2$
$$m = 2a + 2$$
$$= 2(2) + 2$$
$$= 6$$
The point at $x = 2$ is $(2,4)$ since if we plug $x = 2$ into $f(x)$ we end up with $y = 4$
So,
$$y = mx+b$$
$$4 = 6(2) + b$$
$$4 - 12 = b$$
$$-8 = b$$
Therefore your equation is:
$$y = 6x - 8$$
