Question

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

Answer 1

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

Explanation:

Let us consider the equation

`y=-4x^2`

`\mathrm{Domain\:of\:}\:-4x^2\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:<x<\infty \\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:\infty \:\right)\end{bmatrix}`

`\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=(m+n)/(2)`

`y=-4x^2`

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

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

`x_v=(m+n)/(2)`

`x_v=(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<0,\:\mathrm{then\:the\:vertex\:is\:a\:maximum\:value}`

`\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.