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

Jan 21, 2015

You are in the simplest case of this kind of exercises, since you already have the parabola written in the form $y = a {x}^{2} + b x + c$.

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

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

So, $- \setminus \frac{b}{2 a} = - \frac{\setminus \frac{3}{2}}{\setminus \frac{2}{4}} = - \setminus \frac{3}{2} \setminus \frac{4}{2} = 3$
and ${b}^{2} - 4 a c$ equals $\setminus \frac{9}{4} - 4 \setminus \frac{1}{4} \setminus \frac{1}{4} = 2$.

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

The focus of your parabola has thus coordinates $\left(3 , - 1\right)$

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]}