# How do you find the equation of a parabola given the coordinates of the vertex: (2,3) point : (-1,6)?

Apr 23, 2015

Assuming the axis of symmetry is vertical
then the axis of symmetry is $x = 2$
and since $\left(- 1 , 6\right) = \left(2 - 3 , 6\right)$ is a given point on the parabola
then so is $\left(2 + 3 , 6\right) = \left(5 , 6\right)$

The general form of the equation for a parabola (with a vertical axis of symmetry) is
$a {x}^{2} + b x + c = y$

We can substitute our three points that we know into this general form:

$a {\left(- 1\right)}^{2} + b \left(- 1\right) + c = 6$

$a {\left(2\right)}^{2} + b \left(2\right) + c = 3$

$a {\left(5\right)}^{2} + b \left(5\right) + c = 6$

and solve for $a , b , \text{ and } c$

obtaining
$a = \frac{1}{3} , b = - \frac{4}{3} , \text{ and } c = \frac{13}{3}$

So the equation is
$y = \frac{1}{3} {x}^{2} - \frac{4}{3} x + \frac{13}{3}$

Note if the axis of symmetry is horizontal, a similar method can be used; but if the axis of symmetry is other than vertical or horizontal there is insufficient information to derive the parabolic equation.

Apr 23, 2015

Assuming that it is a vertical parabola, the equation looks like:

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

So this parabola has equation:

y=a(x-2)^2+3#

Knowing that $\left(- 1 , 6\right)$ lies on the parabola also tells us that when I put in $- 1$ for $x$ and $6$ for $y$, the equation will be true. That lets us find $a$

$6 = a {\left(- 1 - 2\right)}^{2} + 3$

$6 = 9 a + 3$

$9 a = 3$, so

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

The desired equation is: $y = \frac{1}{3} {\left(x - 2\right)}^{2} + 3$

If the parabola is horizontal, the answer is different.