For questions 1 and 2, use integer values of x from –3 to 3 to graph the equation.

1.  
y = –2x2 + 3  (1 point)

y = -2 x^2 + 3

well that is a parabola.
for big - or + x, it is huge negative so opens down, sheds water.
It is symmetric about x = 0 so that is the vertex (0,3)
Now try your numbers
if x = -3
y = -2(-3)(-3) + 3
y = -18 + 3
y = -15
etc
(same y for x = + 3 of course)

To graph the equation y = -2x^2 + 3, we need to substitute different integer values of x and find the corresponding values of y. Given the range of x from -3 to 3, let's substitute these values one by one:

For x = -3:
y = -2(-3)^2 + 3
y = -2(9) + 3
y = -18 + 3
y = -15

For x = -2:
y = -2(-2)^2 + 3
y = -2(4) + 3
y = -8 + 3
y = -5

For x = -1:
y = -2(-1)^2 + 3
y = -2(1) + 3
y = -2 + 3
y = 1

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

For x = 1:
y = -2(1)^2 + 3
y = -2(1) + 3
y = -2 + 3
y = 1

For x = 2:
y = -2(2)^2 + 3
y = -2(4) + 3
y = -8 + 3
y = -5

For x = 3:
y = -2(3)^2 + 3
y = -2(9) + 3
y = -18 + 3
y = -15

Now, let's plot these points on a graph with x-coordinate on the x-axis and y-coordinate on the y-axis:

(-3, -15)
(-2, -5)
(-1, 1)
(0, 3)
(1, 1)
(2, -5)
(3, -15)

Joining these points, we get a downward-opening parabola.