Question: Find the normal equation to the graph $ y = 2x^2$, that goes through the point $ (1,0.25)$. How this is done? Im not even really sure what I'm being asked here...

Answer: 
The Question
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is to find a line passing through the point $(1,0.25)$ (doesn't lie on the graph) and perpendicular to the tangent line to the $y=2x^2$ at the point of intersection.

Solution
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The required line passing through the point $(1,0.25)$ and a point $(a,2a^2)$ of the graph. The slope of the tangent line at a point $(a,b)=(a,2a^2)$ is $y'(a)=4a.$
Now the slope of the normal is
$$\frac{-1}{4a}=\frac{2a^2-0.25}{a-1}$$
Solving for $a$, we get $a=0.5$ and find the slope is $-0.5$ and the line equation is
$$y-0.25=-0.5(x-1)$$
The Normal is the line passing through a given point which is perpendicular to the tangent line.

So if you can find the equation for the tangent line to the graph $y = 2x^{2}$, and if you know how to find the equation of a line perpendicular to another one, then you should be set.