# How do you write the standard form of the equation of the parabola that has (-1/4, 3/2) vertex and (-2,0) point?

Sep 15, 2016

$y = - \frac{24}{49} {\left(x + \frac{1}{4}\right)}^{2} + \frac{3}{2}$ or $y = \left(- \frac{24}{49}\right) {x}^{2} - \left(\frac{12}{49}\right) x + \frac{72}{49}$

#### Explanation:

Use the equation $y = a {\left(x - h\right)}^{2} + k$ where $\left(h , k\right)$ is the vertex and a is a constant.

The given vertex is $\left(- \frac{1}{4} , \frac{3}{2}\right)$.
Plugging this into the equation above gives

$y = a {\left(x - - \frac{1}{4}\right)}^{2} + \frac{3}{2}$

$y = a {\left(x + \frac{1}{4}\right)}^{2} + \frac{3}{2}$ (EQUATION A)

To find the constant $a$, substitute the given point $\left(- 2 , 0\right)$ in for $x$ and $y$.

$0 = a {\left(- 2 + \frac{1}{4}\right)}^{2} + \frac{3}{2}$

$0 = a {\left(- \frac{7}{4}\right)}^{2} + \frac{3}{2}$
$0 = a \left(\frac{49}{16}\right) + \frac{3}{2}$
$0 - \frac{3}{2} = a \left(\frac{49}{16}\right)$

$\frac{16}{49} \cdot - \frac{3}{2} = a \left(\frac{49}{16}\right) \cdot \frac{16}{49}$
$- \frac{24}{49} = a$

Substitute $a$ into equation A above.
$y = a {\left(x + \frac{1}{4}\right)}^{2} + \frac{3}{2}$
$y = - \frac{24}{49} {\left(x + \frac{1}{4}\right)}^{2} + \frac{3}{2}$

Some books and teachers call this the standard form; but most call it the vertex form. To find the "usual" standard form, square the binomial and simplify.
$y = - \frac{24}{49} \left(x + \frac{1}{4}\right) \left(x + \frac{1}{4}\right) + \frac{3}{2}$
$y = - \frac{24}{49} \left({x}^{2} + \frac{1}{2} x + \frac{1}{16}\right) + \frac{3}{2}$
$y = \left(- \frac{24}{49}\right) {x}^{2} - \left(\frac{12}{49}\right) x - \frac{3}{98} + \frac{3}{2}$
$y = \left(- \frac{24}{49}\right) {x}^{2} - \left(\frac{12}{49}\right) x + \frac{72}{49}$