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

Jul 11, 2018

$y = - 4 {x}^{2}$

#### Explanation:

$\text{the equation of a parabola in "color(blue)"vertex form}$ is.

•color(white)(x)y=a(x-h)^2+k

$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$

$\text{here "(h,k)=(0,0)" thus}$

$y = a {x}^{2}$

$\text{to find a substitute "(-1,-4)" into the equation}$

$- 4 = a$

$y = - 4 {x}^{2} \leftarrow \textcolor{b l u e}{\text{equation of parabola}}$
graph{-4x^2 [-10, 10, -5, 5]}

${x}^{2} = - \frac{1}{4} y \setminus \quad$ or $\setminus \quad {y}^{2} = - 16 x$

#### Explanation:

There are two such parabola satisfying the given conditions as follows

Case 1: Let the vertical parabola with the vertex at $\left(0 , 0\right)$ be

${x}^{2} = k y$

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

${\left(- 1\right)}^{2} = k \left(- 4\right)$

$k = - \frac{1}{4}$

hence setting $k = - \frac{1}{4}$, the equation of vertical parabola

${x}^{2} = - \frac{1}{4} y$

Case 2: Let the horizontal parabola with the vertex at $\left(0 , 0\right)$ be

${y}^{2} = k x$

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

${\left(- 4\right)}^{2} = k \left(- 1\right)$

$k = - 16$

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

${y}^{2} = - 16 x$