represents the x-intercepts and the vertex of the quadratic relation y = −x2 − 6x + 27?

The x-intercepts and the vertex of the quadratic relation y = -x^2 - 6x + 27 can be found by following these steps:

1. Find the x-intercepts: The x-intercepts are the values of x when y = 0. So, we set y = 0 and solve for x:

0 = -x^2 - 6x + 27 This is a quadratic equation in the form ax^2 + bx + c = 0. We can solve it using the quadratic formula x = [-b ± sqrt(b^2 - 4ac)] / (2a):

x = [6 ± sqrt((-6)^2 - 4*(-1)27)] / (2(-1)) x = [6 ± sqrt(36 + 108)] / -2 x = [6 ± sqrt(144)] / -2 x = [6 ± 12] / -2 So, the x-intercepts are x = -3 and x = -9.

2. Find the vertex: 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) = ah^2 + bh + c:

h = -(-6) / (2*(-1)) = 3 k = -3^2 - 6*3 + 27 = -9 - 18 + 27 = 0 So, the vertex is (3, 0).