# How do you write a quadratic equation with Vertex: (-1,-3) x-intercept: 2?

May 25, 2016

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

#### Explanation:

The parabola is of the form:

$y = a {\left(x - h\right)}^{2} + k$

$\text{The vertex will be at the point} \left(\setminus h , \setminus k\right)$

$\text{Given the vertex at } \left(- 1 , - 3\right)$

$h = - 1 \text{ and } k = - 3$

$\text{The " x "- intercept is at " 2 " so the point "(2,0)" is a particular point of the parabola}$

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

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

$0 = a \cdot {\left(3\right)}^{2} - 3$

$9 a - 3 = 0$

$3 a - 1 = 0$

$a = \frac{1}{3}$

$\text{The quadratic equation is: }$

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