y2 = 32x
y = ±√32x
the vertex is (0, 0) and the axis of symmetry is x-axis or y = 0
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I'm assuming that you meant y = 2(x^2) +4. If it were only y = 2x +4, then this would be a linear equation and not a parabola. Anyways, use the equation x = -b/2a to find the x-value of your vertex AND your axis of symmetry. (Given the standard equation y = a(x^2) + bx + c) So, x = -0/2(2) - x = 0 (Axis of Symmetry) Now plug 0 back into your equation to find your y-value of your vertex. y = 2(0^2)+4 y=0 + 4 y = 4 Therefore Vertex = (0,4)
The vertex must be half way between the two x intercepts
x=100
You can find a vertex wherever two lines (or line segments) meet.
4x= y