The x-intercepts of the quadratic equation are found by setting y = 0 and solving for x.

So, we have:

0 = -x^2 - 6x + 27

This can be factored to:

0 = -(x - 3)^2 + 9

Setting each factor equal to zero gives the solutions x = 3.

The vertex of a parabola y = ax^2 + bx + c is given by the point (h, k), where h = -b/2a and k = f(h).

In this case, a = -1 and b = -6, so h = -(-6)/2*(-1) = 3.

Substituting x = 3 into the equation gives:

y = -(3)^2 - 6*3 + 27 = 9

So, the vertex of the parabola is (3, 9).

Therefore, the x-intercepts and the vertex of the quadratic relation y = -x^2 - 6x + 27 are x = 3 and (3, 9) respectively.

Question

The x-intercepts of the quadratic equation are found by setting y = 0 and solving for x.

So, we have:

0 = -x^2 - 6x + 27

This can be factored to:

0 = -(x - 3)^2 + 9

Setting each factor equal to zero gives the solutions x = 3.

The vertex of a parabola y = ax^2 + bx + c is given by the point (h, k), where h = -b/2a and k = f(h).

In this case, a = -1 and b = -6, so h = -(-6)/2*(-1) = 3.

Substituting x = 3 into the equation gives:

y = -(3)^2 - 6*3 + 27 = 9

So, the vertex of the parabola is (3, 9).

Therefore, the x-intercepts and the vertex of the quadratic relation y = -x^2 - 6x + 27 are x = 3 and (3, 9) respectively.

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