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 $x = 2$?

 A: 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:
$$y = 6x - 8$$
A: General formulas
$y=f(x)=x^2+2x-4$
$m=f'(x)=2x+2$  Slope equation
Point-Slope equation of a line
$(y-y_0)=m_0\cdot(x-x_0)$ 
Compute actual values
$x_0 = 2$
$y_0=f(x_0)=f(2)=(2)^2+2(2)-4=4+4-4=4$ 
$m_0=f'(x_0)=f'(2)=2(2)+2=4+2=6$
Substitute values into the line equation
$y-4=6\cdot(x-2)$
$y=6\cdot(x-2)+4$
$y=6x-8$
General Formula for a tangent line to a Parabola @ point $x_0$ on the Parabola
$y=f'(x_0)\cdot(x-x_0)+f(x_0)$
A: Consider the point $(x_1,y_1)=(x_1,f(x_1))$ for the general curve  $y=f(x)$ where tangent equation is desired.
Elementary differentiation familiarity is assumed. The tangent has general equation
$$ \dfrac{y-y_1}{x-x_1}= \dfrac {dy}{dx} \big{|}_{x=x_1.}  $$
For the particular parabola given, plug in coordinate values
$$ \dfrac{y-4}{x-2}= 6  $$
Or,
$$ y=6x-8. $$
A: 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 $y = 6x-8$.
A: An answer that assumes no knowledge of the tangent line equation. 
We are looking for the tangent line to the graph of 
$$f(x)= x²+2x - 4$$ $$at$$ $$x=2$$. 
(1) The tangent line will " touch" the graph of function $f$ at point $( 2,f(2)) = (2,4)$. Let's call this point $P$. 
(2) The equation of the tangent line will have form : $y= ax+b$. ( Any line that is neither horizontal nor vertical has an equation with such a form). So our problem is reduced to : what is number $a$? what is number $b$? 
(3) The slope of the tangent , namely, number " $a$, " will be , by definition, the number $f'(2)$. ( Explanation : the output of the derivative function $f'$ for an arbitrary input  $x = c$ is the instantaneous rate of change of function $f$ at $x=c$, and this instantaneous rate of change of $f$ is nothing else but the slope of the tangent line to the graph of function $f$ at point $( c, f(c) )$  ). 
Since ( assuming  derivation formula : $[ax^n]'= a(n)x^{n-1}$ ) 
$$f'(x)=2x+2$$, 
we consequently have $a = f'(X_P)=f'(2) = 2(2)+2 = 6$. 
(4) Plugging $a=6$ in the equation of the tangent, we get : 
$$y=6x+b$$. 
(5) Any point of the tangent will verify the equation of this line, and, in particular, the point $P = (X_P, Y_P)=(2, 4)$. ( Let's recall that, by definition,  point $P$ lies both on the graph of function $f$ and on the tangent line to this graph at $x=2$). Since, as we said, point P " obeys" the equation of the tangent, we have :  
$$Y_P=6X_P+b$$
But we know that : $Y_P=4$ and $X_P=2$
Therefore , 
$4=6(2)+b$ , which implies $4 = 12 + b$ and $b=-8$. 
(6) Having found both the slope ( number $a$ ) and number $b$, we can write that the equation of the tangent line we were looking for is : 
$$y=6x + (-8)$$
or 
$$y=6x-8$$. 
