Equation of line normal to $y = x^3 -2x^2$ at $x=0$ 
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*Find the equation of normal to the curve $y =x^3 - 2x^2$ at $x= 0.$

*Find the co-ordinates of the point of intersection of the normal and
the line $y = 4.$


I differentiated the equation with respect $x$ and I got: $dy/dx=3x^2-4x$. But the slope of normal should be a non-zero, finite term right? I just don't know how to proceed further with this. I tried double-differentiating it but I'm pretty sure that's not the correct method.
 A: Well, sort of. 
Slope can indeed be zero: that signifies that the line tangent to the curve at $x=0$, after evaluating $dy/dx$ at $0$ gives $\dfrac{dy}{dx}= 0$. 
When slope equals zero, we know we have a horizontal line tangent at $x=0$, in this case. That horizontal line is given by $y(0) = 0$.
Since the line tangent at $0$ is horizontal, the normal at $x = 0$ will be a vertical line (slope is undefined), indeed, it is given by the line $x=0
The point of intersection between $x=0$ and $y = 4$ is precisely $(0, 4)$.
A: If you have the derivative, you have the slope of the tangent line. You know that the normal must be perpendicular to this tangent line. 
If you have a slope of $0$ for your tangent line, what's the slope (crudely speaking) of your normal? A negative reciprocal... which is more or less an infinitely steep line; that is, a straight vertical line. Thus, the solution is $x=0$. 
A: The first helper to you is the plot of this function:
http://www.wolframalpha.com/input/?i=y+%3Dx%5E3+-+2x%5E2
So you can easily see that the normal is vertical in the $x=0$. So the intersection with the line $y=4$ comes at the point $(0,4)$.
