# How do you write the quadratic equation given Vertex: (-2,0) Passing through: (0,3)?

$y = \frac{3}{4} {x}^{2} + 3 x + 3$

#### Explanation:

From the given vertex $\left(h , k\right) = \left(- 2 , 0\right)$ and passing thru $\left(0 , 3\right)$

A little inspection tells us that the vertex is lower than the given point. We can conclude that the parabola opens upward.

${\left(x - h\right)}^{2} = + 4 p \left(y - k\right)$

Let us use the two given points to solve for p:

${\left(x - h\right)}^{2} = + 4 p \left(y - k\right)$
${\left(0 - - 2\right)}^{2} = + 4 p \left(3 - 0\right)$
$4 = 4 p \cdot 3$

$4 = 12 p$
$p = \frac{1}{3}$

Now we can write the equation
${\left(x - h\right)}^{2} = + 4 p \left(y - k\right)$
${\left(x - - 2\right)}^{2} = 4 \left(\frac{1}{3}\right) \left(y - 0\right)$
$y = \frac{3}{4} \left({x}^{2} + 4 x + 4\right)$

$y = \frac{3}{4} {x}^{2} + 3 x + 3$

God bless....I hope the explanation is useful.