Question 420127
When you compare the given equation: {{{f(x) = -2x^2}}} with the general form for a quadratic equation: {{{f(x) = ax^2+bx+c}}}, you can see that: a = -2, b = 0, and c = 0.
The x-coordinate of the vertex is given by:
{{{x = (-b)/2a}}} and, since b= 0, the x-coordinate is x = 0.
Substitute x = 0 into the given equation: {{{y = -2x^2}}} to get:
{{{y = -2(0)^2}}}
{{{y = 0}}}
The vertex is at (0, 0)
The graph of the given equation is a parabola that opens downward (negative coefficient of the {{{x^2}}} term.
The equation of the axis of symmetry is {{{x = (-b)/2a}}} or {{{x = 0}}} which is the y-axis.
Here's the graph of the given equation:
{{{graph(400,400,-5,5,-5,5,-2x^2)}}}