What are the axes of symmetry in the graphs of the equations -2x^2 and -2x^2 + 4? I think the vertices are (0,0) and (0,4).

You talk about equations, but show no equation.

You must have meant the equations to be
y = -2x^2 and y = -2x^2 + 4

You are asking for the axes of symmetry but instead state the vertices.
You were correct that the vertices are (0,0) and (0,4) respectively.

so the axes of symmetry are x = 0 for both of them,
that is, the axis of symmetry is the y-axis

Thank you. Yes, I meant to put the y in there. Are the x-intercepts 0 and 4, respectively?

I mean 0 for -2x^2. Sorry about the 4, I didn't mean to put that.

yes , the x-intercepts are 0 and 4, respectively

confused about your second reply, the vertices are indeed (0,0) and (0,4)

<<I mean 0 for -2x^2. Sorry about the 4, I didn't mean to put that. >> , don't even know what that is supposed to say.

see,
http://www.wolframalpha.com/input/?i=y+%3D+-2x%5E2+%2C+y+%3D+-2x%5E2+%2B+4

To determine the axes of symmetry for the graphs of the equations -2x^2 and -2x^2 + 4, we need to use the fact that the equations are functions of x.

The equation -2x^2 represents a downward facing parabola since the coefficient of x^2 is negative. The vertex of this parabola can be found using the formula x = -b/2a, where a and b are the coefficients of x^2 and x, respectively. In this case, a = -2 and b = 0. Therefore, the x-coordinate of the vertex is given by x = 0 / (2 * (-2)) = 0. This means that the vertex is (0, 0).

Similarly, for the equation -2x^2 + 4, the vertex can be found by considering the x-coordinate to be x = -b/2a. In this case, a = -2 and b = 0. Thus, the x-coordinate of the vertex is x = 0 / (2 * (-2)) = 0, which gives the same vertex as before: (0, 0).

Therefore, the equations -2x^2 and -2x^2 + 4 have the same vertex, which is (0, 0). Since the vertex lies on the axis of symmetry, the axis of symmetry for both equations is the vertical axis x = 0.