Question

GRAPH the parabola.y= -3x^2+3Plot five points on the parabola: the vertex, two points to the left of the vertex, and two points to the right of the vertex.

Answer 1

Okay, here we have this;

Considering the provided parabola, we are going to calculate the requested vertex and two points on each side of the vertex to later graph it, then we have:

Vertex:

Recall that in an equation of the form: y=a(x−h)^2+k. the vertex of the parabola is the point (h,k).

So according to that, in this case for our function we have:

y=a(x−h)^2+k

y= -3(x-0)^2+3

Then, the vertex of the function is (0,3)

Now we will calculate two points to the left taking x=-1 and x=-2, and two to the right taking x=1, x=2.

x=-1:

y= -3x^2+3

y= -3(-1)^2+3

y= -3(1)+3

y= -3+3

y=0

x=-2:

y= -3x^2+3

y= -3(-2)^2+3

y= -3(4)+3

y= -12+3

y=-9

x=1:

y= -3x^2+3

y= -3(1)^2+3

y= -3(1)+3

y= -3+3

y=0

x=2:

y= -3x^2+3

y= -3(2)^2+3

y= -3(4)+3

y= -12+3

y=-9

Now plotting the points we have: