What is the equation of the parabola that has a vertex at  (2, -1)  and passes through point  (3,-4) ?

Apr 17, 2018

$y = - 3 {x}^{2} + 12 x - 13$

Explanation:

This problem will be easier if we express the equation for the parabola in vertex form.

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

where $k$ is the $x$-coordinate of the vertex and $h$ is the $y$-coordinate of the vertex. Since the vertex is at $\left(2 , - 1\right)$ our equation for the parabola becomes

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

Since the parabola passes through the point $\left(3 , - 4\right)$ we can write

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

and solve for $a$ by adding 1 to both sides.

$a = - 3$

So the equation for the parabola is

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

We can expand the binomial to obtain the equation for the parabola in standard form.

$y = - 3 \left({x}^{2} - 4 x + 4\right) - 1 = - 3 {x}^{2} + 12 x - 13$

graph{-3x^2+12x-13 [-1, 4, -10, 5]}