How do you graph the parabola #y=2x^2-2# using vertex, intercepts and additional points?
1 Answer
Jan 20, 2018
Vertex (0,-2)
y-intercept: (0,-2)
x-intercepts: (-1,0), (1,0)
additional points: (-2,6), (2,6)
Explanation:
Your parabola is already as simple as it can be.
The vertex is (0,-2).
Therefore, your y-intercept is (0,-2).
Your x-intercept is (-1,0) and (1,0)
You get this by setting
You get
For additional points, you can plug in 2 and -2, which you get 6 for both because it is an even function.
