y=–x2+12x–4

1 answer

The given equation is a quadratic equation in the standard form.

To analyze this equation, we can first determine the vertex of the parabola represented by the equation.

The x-coordinate of the vertex is given by -b/2a, where a and b are the coefficients of the quadratic equation. In this case, a = -1 and b = 12.

x-coordinate of the vertex = -12/ (2 * -1) = -12 / -2 = 6

To find the y-coordinate of the vertex, we substitute the x-coordinate into the equation.

y = -6^2 + 12(6) - 4 = -36 + 72 - 4 = 32

Therefore, the vertex of the parabola is (6, 32).

The parabola opens downwards because the coefficient of the x^2 term is negative. The vertex is the highest point.

Now, we can determine the axis of symmetry, which is a vertical line passing through the vertex. The equation of the axis of symmetry is x = 6.

Finally, we can find the y-intercept by substituting x = 0 into the equation.

y = -0^2 + 12(0) - 4 = 0 + 0 - 4 = -4

Therefore, the y-intercept is -4.

In summary, the vertex of the parabola is (6, 32), the axis of symmetry is x = 6, and the y-intercept is -4.