Find the x-intercept(s) and the coordinates of the vertex for the parabola =y−x2−2x15. If there is more than one x-intercept, separate them with commas.

The equation you've given seems to be a bit unclear. However, if the equation of the parabola is y = -x^2 - 2x + 15, we can proceed as follows:

1. Find the x-intercepts: The x-intercepts of a parabola are the x-values where y = 0. So, we set y = 0 in the equation and solve for x:

0 = -x^2 - 2x + 15 => x^2 + 2x - 15 = 0

This is a quadratic equation in the form ax^2 + bx + c = 0, where a = 1, b = 2, and c = -15. We can solve for x using the quadratic formula x = [-b ± sqrt(b^2 - 4ac)] / (2a):

x = [-2 ± sqrt((2)^2 - 41(-15))] / (2*1) => x = [-2 ± sqrt(4 + 60)] / 2 => x = [-2 ± sqrt(64)] / 2 => x = [-2 ± 8] / 2 => x = 3, -5

So, the x-intercepts of the parabola are x = 3 and x = -5.

2. Find the coordinates of 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 = y(h).

h = -(-2) / (21) = 1 k = -(1)^2 - 21 + 15 = -1 - 2 + 15 = 12

So, the vertex of the parabola is at (1, 12).