Question

Which of the equations below could be the equation of this parabola?

Answer 1

Answer:

 $y=-4x^2$  is the equation of this parabola.

Step-by-step explanation:

Let us consider the equation

$y=-4x^2$

$\mathrm{Domain\:of\:}\:-4x^2\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:$

$\mathrm{Range\:of\:}-4x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le \:0\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:0]\end{bmatrix}$

$\mathrm{Axis\:interception\:points\:of}\:-4x^2:\quad \mathrm{X\:Intercepts}:\:\left(0,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:0\right)$

As

$\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=a\left(x-m\right)\left(x-n\right)$

$\mathrm{is\:the\:average\:of\:the\:zeros}\:x_v=\frac{m+n}{2}$

$y=-4x^2$

$\mathrm{The\:parabola\:params\:are:}$

$a=-4,\:m=0,\:n=0$

$x_v=\frac{m+n}{2}$

$x_v=\frac{0+0}{2}$

$x_v=0$

$\mathrm{Plug\:in}\:\:x_v=0\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}$

$y_v=-4\cdot \:0^2$

$y_v=0$

Therefore, the parabola vertex is

$\left(0,\:0\right)$

$\mathrm{If}\:a$

$\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}$

$a=-4$

$\mathrm{Maximum}\space\left(0,\:0\right)$

so,

$\mathrm{Vertex\:of}\:-4x^2:\quad \mathrm{Maximum}\space\left(0,\:0\right)$

Therefore,  $y=-4x^2$  is the equation of this parabola. The graph is also attached.