Question

I have nooooo clue, please help

Answer 1

Answer:

case a) $x^{2}=3y$ ----> open up

case b) $x^{2}=-10y$ ----> open down

case c) $y^{2}=-2x$ ----> open left

case d) $y^{2}=6x$ ----> open right

Step-by-step explanation:

we know that

1) The general equation of a vertical parabola is equal to

$y=a(x-h)^{2}+k$

where

a is a coefficient

(h,k) is the vertex

If a>0 ----> the parabola open upward and the vertex is a minimum

If a<0 ----> the parabola open downward and the vertex is a maximum

2) The general equation of a horizontal parabola is equal to

$x=a(y-k)^{2}+h$

where

a is a coefficient

(h,k) is the vertex

If a>0 ----> the parabola open to the right

If a<0 ----> the parabola open to the left

Verify each case

case a) we have

$x^{2}=3y$

so

$y=(1/3)x^{2}$

$a=(1/3)$

so

$a>0$

therefore

The parabola open up

case b) we have

$x^{2}=-10y$

so

$y=-(1/10)x^{2}$

$a=-(1/10)$

$a<0$

therefore

The parabola open down

case c) we have

$y^{2}=-2x$

so

$x=-(1/2)y^{2}$

$a=-(1/2)$

$a<0$

therefore

The parabola open to the left

case d) we have

$y^{2}=6x$

so

$x=(1/6)y^{2}$

$a=(1/6)$

$a>0$

therefore

The parabola open to the right