How do you find the vertex and axis of symmetry for #y = -x^2 - 10x#?

The equation must be rewritten so that the x terms form a perfect square in order to determine the vertex and the axis of symmetry:

This results in x=-5 for the axis of symmetry and (-5,25) for the vertex.

To find the vertex of the parabola represented by the equation y = -x^2 - 10x, first, complete the square to rewrite the equation in vertex form. Then, identify the vertex and axis of symmetry.

Completing the square: y = -x^2 - 10x = -(x^2 + 10x)

To complete the square, we need to add and subtract (10/2)^2 = 25 inside the parentheses: = -(x^2 + 10x + 25 - 25)

Now, rewrite it: = -(x^2 + 10x + 25) + 25

Factor the quadratic expression inside the parentheses: = -(x + 5)^2 + 25

Now, the equation is in vertex form, y = a(x - h)^2 + k, where (h, k) is the vertex. Therefore, the vertex is (-5, 25).

The axis of symmetry is a vertical line passing through the vertex. Since the x-coordinate of the vertex is -5, the equation of the axis of symmetry is x = -5.