Answer:
[tex]y=-4x^2[/tex] is the equation of this parabola.
Step-by-step explanation:
Let us consider the equation
[tex]y=-4x^2[/tex]
[tex]\mathrm{Domain\:of\:}\:-4x^2\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:<x<\infty \\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:\infty \:\right)\end{bmatrix}[/tex]
[tex]\mathrm{Range\:of\:}-4x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le \:0\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:0]\end{bmatrix}[/tex]
[tex]\mathrm{Axis\:interception\:points\:of}\:-4x^2:\quad \mathrm{X\:Intercepts}:\:\left(0,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:0\right)[/tex]
As
[tex]\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=a\left(x-m\right)\left(x-n\right)[/tex]
[tex]\mathrm{is\:the\:average\:of\:the\:zeros}\:x_v=\frac{m+n}{2}[/tex]
[tex]y=-4x^2[/tex]
[tex]\mathrm{The\:parabola\:params\:are:}[/tex]
[tex]a=-4,\:m=0,\:n=0[/tex]
[tex]x_v=\frac{m+n}{2}[/tex]
[tex]x_v=\frac{0+0}{2}[/tex]
[tex]x_v=0[/tex]
[tex]\mathrm{Plug\:in}\:\:x_v=0\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}[/tex]
[tex]y_v=-4\cdot \:0^2[/tex]
[tex]y_v=0[/tex]
Therefore, the parabola vertex is
[tex]\left(0,\:0\right)[/tex]
[tex]\mathrm{If}\:a<0,\:\mathrm{then\:the\:vertex\:is\:a\:maximum\:value}[/tex]
[tex]\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}[/tex]
[tex]a=-4[/tex]
[tex]\mathrm{Maximum}\space\left(0,\:0\right)[/tex]
so,
[tex]\mathrm{Vertex\:of}\:-4x^2:\quad \mathrm{Maximum}\space\left(0,\:0\right)[/tex]
Therefore, [tex]y=-4x^2[/tex] is the equation of this parabola. The graph is also attached.
