The equation you provided seems to be a quadratic function in the form of y = ax^2 + b. The axis of symmetry for a parabola in this form is given by the formula x = -b/2a.

However, in your equation, b = 12 and a = 4, which doesn't seem correct as 'b' is usually the coefficient of 'x' and 'a' is the coefficient of 'x^2'.

If your equation is y = 4x^2 + 12, then 'b' is actually 0 (since there is no 'x' term) and 'a' is 4.

In this case, the axis of symmetry would be x = -0/(2*4) = 0.

So, the axis of symmetry of the parabola y = 4x^2 + 12 is x = 0.

Question

The equation you provided seems to be a quadratic function in the form of y = ax^2 + b. The axis of symmetry for a parabola in this form is given by the formula x = -b/2a.

However, in your equation, b = 12 and a = 4, which doesn't seem correct as 'b' is usually the coefficient of 'x' and 'a' is the coefficient of 'x^2'.

If your equation is y = 4x^2 + 12, then 'b' is actually 0 (since there is no 'x' term) and 'a' is 4.

In this case, the axis of symmetry would be x = -0/(2*4) = 0.

So, the axis of symmetry of the parabola y = 4x^2 + 12 is x = 0.

Knowee AI · Accepted Answer

The equation you provided seems to be a quadratic function in the form of y = ax^2 + b. The axis of symmetry for a parabola in this form is given by the formula x = -b/2a.

However, in your equation, b = 12 and a = 4, which doesn't seem correct as 'b' is usually the coefficient of 'x' and 'a' is the coefficient of 'x^2'.

If your equation is y = 4x^2 + 12, then 'b' is actually 0 (since there is no 'x' term) and 'a' is 4.

In this case, the axis of symmetry would be x = -0/(2*4) = 0.

So, the axis of symmetry of the parabola y = 4x^2 + 12 is x = 0.

Find the equation of the axis of symmetry of the following parabola algebraically.y, equals, 4, x, squared, plus, 12y=4x 2 +12

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