Question 422555
{{{y+4x^2=0}}}


{{{y= -4x^2}}}........compare with the general quadratic equation

{{{y = ax^2 + bx + c}}}

Here, {{{a = -4 }}}, {{{b =0}}}, and {{{c = 0}}}

Substitute {{{a = -4 }}} and {{{b =0}}} in the equation of axis of symmetry,

{{{x= -b/2a}}}

{{{x=-0/2(-4)}}}

{{{x=0}}}


so, the equation of the axis of symmetry for this parabola is {{{x = 0}}}

Since the equation for the axis of symmetry is {{{x = 0}} and the {{{vertex}}} lies on the axis, the {{{x–coordinate}}} for the vertex is {{{0}}}.

Substitute {{{x = 0}}} in the given equation {{{y= -4x^2}}}

{{{y= -4*0^2}}}

{{{y = 0}}}

The coordinates of {{{vertex}}} are (0, 0).

The coefficient of {{{x^2}}} is negative, therefore the vertex is a maximum and
the graph of the equation opens down.

{{{ graph( 500, 500, -10, 10, -10, 10, -4x^2) }}}