How do I find the equation of the normal to $y = x^2 - 3x +2$ The question I got first asked me to solve the simultaneous equation:
$y = x^2 - 3x +2$ and
$y = 3x - 7$ 

then hence or otherwise find the equation of the normal of $y = x^2 - 3x +2$ at one point $(3,2)$. Giving the answer in the form of $ax + by + c =0$

How do I solve this?
When solving the system of equations, you'll obtain a quadratic with exactly one real solution at $x = 3$. This implies that the given line is actually tangent to the given parabola.

Coincidentally, the second part of the question asks for the normal line at $x = 3$. Since perpendicular lines have slopes that multiply to be $-1$, and the slope of the tangent line is $3$, it follows that the slope of the normal line is its negative reciprocal, $-1/3$. Using the point-slope formula, we have:
$$
y - 2 = \frac{-1}{3}(x - 3)
$$
It remains to convert this to the form $ax + by + c = 0$ by clearing the fraction and bringing all terms to one side.
I assume that ''2Y'' part is a typo? Solving that equation is as follows:


x^2-6x+9=0 
(x-3)^2 = 0
x=3


For the hence/otherwise part


dy/dx = 2x - 3
Substitute in the x co-ordinate to get the gradient of it at the specified point. (which is x = 3)
M(Tangent) = 3, therefore, M(Normal) = -1/3
y= -1/3x + c, sub in point
c= 3 
Therefore -> y= -1/3x + 3 (Multiply both sides by 3)
3y= -x + 9
3y + x - 9 = 0 (as specified in the question)


Therefore, equation to the normal is 3y + x - 9 = 0