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

2 Answers
Apr 15, 2015

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

#y=-3(x+2)^2+3# has #h = -2# , #k = 3# and #a = -3#

so the vertex is #(-2, 3)# 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(x - h)^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(x - h)^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:
myscreenshot