The x-intercepts of the parabola defined by y = x^2 - 9 can be found by setting y to 0 and solving for x:

0 = x^2 - 9
x^2 = 9
x = ± √9
x = ± 3

So, the x-intercepts are x = -3 and x = 3.

The vertex of a parabola defined by y = ax^2 + bx + c is given by the point (h, k), where h = -b/2a and k = c - b^2/4a. In this case, a = 1, b = 0, and c = -9, so:

h = -b/2a = -0/(2*1) = 0
k = c - b^2/4a = -9 - (0)^2/(4*1) = -9

So, the vertex of the parabola is at (0, -9).

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The x-intercepts of the parabola defined by y = x^2 - 9 can be found by setting y to 0 and solving for x:

0 = x^2 - 9
x^2 = 9
x = ± √9
x = ± 3

So, the x-intercepts are x = -3 and x = 3.

The vertex of a parabola defined by y = ax^2 + bx + c is given by the point (h, k), where h = -b/2a and k = c - b^2/4a. In this case, a = 1, b = 0, and c = -9, so:

h = -b/2a = -0/(2*1) = 0
k = c - b^2/4a = -9 - (0)^2/(4*1) = -9

So, the vertex of the parabola is at (0, -9).

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The x-intercepts of the parabola defined by y = x^2 - 9 can be found by setting y to 0 and solving for x:

0 = x^2 - 9
x^2 = 9
x = ± √9
x = ± 3

So, the x-intercepts are x = -3 and x = 3.

The vertex of a parabola defined by y = ax^2 + bx + c is given by the point (h, k), where h = -b/2a and k = c - b^2/4a. In this case, a = 1, b = 0, and c = -9, so:

h = -b/2a = -0/(2*1) = 0
k = c - b^2/4a = -9 - (0)^2/(4*1) = -9

So, the vertex of the parabola is at (0, -9).

the x-intercepts and the vertex of the parabola defined by y = x2 - 9?

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