Question

How do I find the focus of the parabola with the equation `y=1/4x^2-3/2x+1/4`?

Answer 1

You are in the simplest case of this kind of exercises, since you already have the parabola written in the form `y=ax^2+bx+c`.

The general rule says that, in this case, the coordinates of the focus are
`(-\frac{b}{2a}, \frac{1-\Delta}{4a})`,
where `\Delta` is the discriminant `b^2-4ac`.

Let's compute this out for your values: you have `a=\frac{1}{4}`, `b=-\frac{3}{2}`, and `c=a=\frac{1}{4}`.

So, `-\frac{b}{2a}=-frac{\frac{3}{2}}{\frac{2}{4}]=-\frac{3}{2}\frac{4}{2}=3`
and `b^2-4ac` equals `\frac{9}{4}-4\frac{1}{4}\frac{1}{4}=2`.

Thus, `\frac{1-\Delta}{4a}` equals `\frac{1-2}{4\frac{1}{4}}`, namely `-1`.

The focus of your parabola has thus coordinates `(3,-1)`

This is the graph of your parabola, you'll see that the result we got is quite reasonable
graph{x^2/4-3x/2+1/4 [-10, 10, -5, 5]}