# How do write standard form of equation of the parabola given that vertex is (5/2, - 3/4) and graph passes through point (-2 ,4)?

Aug 23, 2016

Express the parabola in vertex form and work from there to obtain $y = \frac{19}{81} {x}^{2} - \frac{95}{81} x + \frac{58}{81}$.

#### Explanation:

We might just want to jump into the problem and try finding the standard form $y = a {x}^{2} + b x + c$ of this parabola, but we don't have enough information to do this yet. Instead, we have to use the vertex form of a parabola:
$y = a {\left(x - h\right)}^{2} + k$
Where $\left(h , k\right)$ is the vertex.

The vertex is given in the problem, so:
$y = a {\left(x - \frac{5}{2}\right)}^{2} - \frac{3}{4}$

We have to find the value of the constant $a$ (which determines how thin or thick the parabola is), and we can do this using the point $\left(- 2 , 4\right)$:
$y = a {\left(x - \frac{5}{2}\right)}^{2} - \frac{3}{4}$
$\to 4 = a {\left(- 2 - \frac{5}{2}\right)}^{2} - \frac{3}{4}$
$\to 4 + \frac{3}{4} = a {\left(- \frac{9}{2}\right)}^{2}$
$\to \frac{19}{4} = \frac{81}{4} a$
$\to a = \frac{19}{81}$

Our equation in vertex form is therefore:
$y = \frac{19}{81} {\left(x - \frac{5}{2}\right)}^{2} - \frac{3}{4}$

We can expand the right hand side to find the standard form:
$y = \frac{19}{81} {\left(x - \frac{5}{2}\right)}^{2} - \frac{3}{4}$
$\to y = \frac{19}{81} \left({x}^{2} - 5 x + \frac{25}{4}\right) - \frac{3}{4}$
$\to y = \frac{19}{81} \left({x}^{2} - 5 x + \frac{25}{4}\right) - \frac{3}{4}$
$\to y = \frac{19}{81} {x}^{2} - \frac{95}{81} x + \frac{475}{324} - \frac{3}{4}$
$\to y = \frac{19}{81} {x}^{2} - \frac{95}{81} x + \frac{58}{81}$