+24 ok:
state the axis of symmetry of the parabola: y=3(x-2)^2-4
also state the vertex, and whether it is a maximum or a minimum, and the y intercept of the parabola and the graph
+129852 y = 3(x - 2)^2 - 4
Vertex = (2, - 4)
Axis of symmetry : x = 2
This parabola opens upward....therefore...the vertex is a minimum
y intercept : let x = 0 and we have
y = 3(0-2)^2 - 4 = 3(-2)^2 - 4 = 3(4) - 4 = 12 - 4 = 10
Here's the graph : https://www.desmos.com/calculator/w1tqaeu5yv
+129852 y = 3( - 2)^2 + ( - 4 )
We have the form
y = a(x - h)^2 + k
When "a" is positive, the parabola opens upward
The vertex = (h, k) = (2, -4 )
The axis of symmery is x = h
The y intercept is found just as I described
The vertex is always the minimum when a parabola opens upward
