Question

What is the equation of the parabola that has a vertex at `(0, 0)` and passes through point `(-1,-4)`?

Answer 1

`y=-4x^2`

Explanation:

`"the equation of a parabola in "color(blue)"vertex form"` is.

`•color(white)(x)y=a(x-h)^2+k`

`"where "(h,k)" are the coordinates of the vertex and a"`
`"is a multiplier"`

`"here "(h,k)=(0,0)" thus"`

`y=ax^2`

`"to find a substitute "(-1,-4)" into the equation"`

`-4=a`

`y=-4x^2larrcolor(blue)"equation of parabola"`
graph{-4x^2 [-10, 10, -5, 5]}

Answer 2

`x^2=-1/4y\quad` or `\quad y^2=-16x`

Explanation:

There are two such parabola satisfying the given conditions as follows

Case 1: Let the vertical parabola with the vertex at `(0, 0)` be

`x^2=ky`

since, above parabola passes through the point `(-1, -4)` then it will satisfy the above equation as follows

`(-1)^2=k(-4)`

`k=-1/4`

hence setting `k=-1/4`, the equation of vertical parabola

`x^2=-1/4y`

Case 2: Let the horizontal parabola with the vertex at `(0, 0)` be

`y^2=kx`

since, above parabola passes through the point `(-1, -4)` then it will satisfy the above equation as follows

`(-4)^2=k(-1)`

`k=-16`

Now, setting `k=-16`, the equation of vertical parabola

`y^2=-16x`