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maks197457 [2]
3 years ago
12

I have nooooo clue, please help

Mathematics
1 answer:
Aleksandr-060686 [28]3 years ago
3 0

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

therefore

The parabola open down

case c) we have

y^{2}=-2x

so

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

a=-(1/2)

a

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

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