what are five points that i can graph for the equation y=x^2+4 (vertex form)
The vertex form is
y = 1(x-0)²+4
y = a(x-h)²+k, so the vertex is (h,k) = (0.4)
(-3,13), (-2,8), (-1,5), (0,4), (1,5), (2,8), (3,13)
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and for the equation
y=2x^2+6x=12
You mistyped an = sign for either a + or a - sign
If you meant a + sign, y=2x²+6x+12
(-4,20), (-3,12), (-2,8), (-1.5,7.5), (-1,-8), (0,12), (1,20)
If you meant a - sign, y=2x²+6x-12
(-5,8), (-4,-4), (-3,-12), (-2,-16), (-1.5,-16.5), (-1,-16),
(0,-12), (1,-4), (2,8)
Edwin
