Figuring out an equation for a tangent line I have a small question, the task is finding an equation for the tangent line at  $g(x)=2x^3+5x^2$ when $x=-2$.
My question is, instead of using calculus, what stops us from just plugging in a $x=3$ and go old fashion $y-y_1=m(x-x_1)$? Since we have the formula, why would we need to further complicate it? Or am I missing something? 
Thanks in advance!
 A: You certainly can't do this.
Yes, the slope of line can be found with two points, but you can't go tossing random numbers into a function to get a tangent line. The reason why we need calculus is the notion of limit: we need two points close to each other so that they're pretty much the same point - hence giving a tangent line.
A: You could ask for the equation of a line through the point $(-2,4)$ which intersects $g(x)$ with multiplicity greater than $1$.  For the line we take the equation to be $y-4=m(x+2)$, so $y=mx+2m+4$.  Then
\begin{align*}
2x^3+5x^2 &= mx+2m+4 \text{ so}\\ 
2x^3+5x^2 - mx - 2m- 4 &= 0
\end{align*}
But we know $x=-2$ must be a root of the cubic, and after dividing by $x+2$ we get the quadratic
$$2x^2+x-(m+2)$$
If the intersection multiplicity is to be greater than $1$, the quadratic also needs to be divisible by $x+2$  The remainder after performing polynomial long division is $-m-2+6$.  For the remainder to be zero as we want, we must have $m=4$.  That gives $y=4x+12$ as the equation of the line.
Comparing what we would find using calculus, $g'(x)=6x^2+10$ giving $g'(-2)= 4$ for the slope of the tangent line at $x=-2$ which actually seems a bit less complicated, no?
A: One way to get an equation of a line is with a point and a slope. This is the appropriate way with the equation you give: $y-y_1 = m(x-x_1)$. You indicate the easy way to get the point yourself, plug in $x=3$. All that remains is the slope, $m$. This is where calculus kicks in. Easiest way to get the slope at $x=3$ is to take the derivative and plug in 3.
$$g'(x)=6x^2+10x$$
$$g'(3)=84$$
So $m=84$ at $x=3$.
