# What is the equation of the parabola that has a vertex at  (-4, 4)  and passes through point  (6,104) ?

Mar 4, 2016

$y = {\left(x + 4\right)}^{2} + 4$ or
$y = {x}^{2} + 8 \cdot x + 20$

#### Explanation:

$y = a \cdot {\left(x - {x}_{v e r t e x}\right)}^{2} + {y}_{v e r t e x}$.

We have $\left(- 4 , 4\right)$ as our vertex, so right off the bat we have

$y = a \cdot {\left(x - \left(- 4\right)\right)}^{2} + 4$ or

$y = a \cdot {\left(x + 4\right)}^{2} + 4$, less formally.

Now we just need to find "$a$."
To do this we sub in the values for the second point $\left(6 , 104\right)$ into the equation and solve for $a$.
Subbing in we find
$\left(104\right) = a \cdot {\left(\left(6\right) + 4\right)}^{2} + 4$
or
$104 = a \cdot {\left(10\right)}^{2} + 4$.

Squaring $10$ and subtracting $4$ from both sides leaves us with
$100 = a \cdot 100$ or $a = 1$.

Thus the formula is $y = {\left(x + 4\right)}^{2} + 4$.

If we want this in standard form ( $y = a \cdot {x}^{2} + b \cdot x + c$ ) we expand the square term to get

$y = \left({x}^{2} + 8 \cdot x + 16\right) + 4$ or

$y = {x}^{2} + 8 \cdot x + 20$.