# What is the vertex and direction of this parabola y=-3(x+2)^2+3?

Apr 15, 2015

$y = a {\left(x - h\right)}^{2} + k$ has
vertex $\left(h , k\right)$ and
opens up if $a$ is positive (also written "if $a > 0$")
opens down if $a$ is negative ("if $a < 0$")

$y = - 3 {\left(x + 2\right)}^{2} + 3$ has $h = - 2$ , $k = 3$ and $a = - 3$

so the vertex is $\left(- 2 , 3\right)$ and the parabola opens down.

Here's the graph:

graph{y = -3(x+2)^2+3 [-12.31, 10.2, -5.625, 5.625]}

Apr 15, 2015

A parabola in the form: $y = a {\left(x - h\right)}^{2} + k$ will open up if a > 0, and down if a < 0. Since a = -3 in your example, the parabola will open down and have a maximum at the vertex.

In the form $y = a {\left(x - h\right)}^{2} + k$, (h,k) is the vertex. In your example, then, (-2,3) is the vertex. The maximum value is 3. Observe the graph below: