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Solve. y=2x^2+8 for the axis of symmetry and the vertex

User Bince
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The equation y = 2x^2 + 8 is in the form of y = ax^2 + bx + c, where a = 2, b = 0, and c = 8.

The axis of symmetry is given by the formula:
x = -b / 2a

Substituting the values of a and b into the formula, we get:
x = -0 / 4 = 0

Therefore, the axis of symmetry is x = 0.

To find the vertex, we substitute the value of x = 0 into the equation:
y = 2x^2 + 8
y = 2(0)^2 + 8
y = 8

Therefore, the vertex is at the point (0, 8).
User MadhavanRP
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