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Arrange the parabolas represented by the equations in order of the positions of their directrixes on the coordinate plane, starting with the lowest and ending with the highest.

BEFORE YOU CLICK THIS I have asked for help with this like 5 times and people have-example-1

2 Answers

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Answer:

The order of equation is


y=(1)/(8)x^2-(1)/(2)x-(1)/(2)


y=-(1)/(12)x^2-(7)/(6)x-(109)/(12)


y=-(1)/(32)x^2-(1)/(16)x-(289)/(32)


y=(1)/(20)x^2+(2)/(5)x+(29)/(5)


y=-(1)/(16)x^2+(3)/(4)x-(21)/(4)


y=(1)/(8)x^2+(3)/(4)x+(41)/(8)


y=-(1)/(52)x^2-(4)/(13)x-(146)/(13)


y=(1)/(28)x^2-(5)/(14)x+(333)/(28)

Explanation:

If a parabola is defined as


y=ax^2+bx+c

then the directrix of the parabola is


y=-(b^2)/(4a)-(1)/(4a)+c

1.

The equation of the parabola is


y=(1)/(8)x^2+(3)/(4)x+(41)/(8)

The directrix of the parabola is


y=-(((3)/(4))^2)/(4((1)/(8)))-(1)/(4((1)/(8)))+((41)/(8))


y=2

Similarly find the directrix of each parabola.

2.

The equation of the parabola is


y=(1)/(20)x^2+(2)/(5)x+(29)/(5)

The directrix of the parabola is
y=0

3.

The equation of the parabola is


y=-(1)/(52)x^2-(4)/(13)x-(146)/(13)

The directrix of the parabola is
y=3

4.

The equation of the parabola is


y=-(1)/(16)x^2+(3)/(4)x-(21)/(4)

The directrix of the parabola is
y=1

5.

The equation of the parabola is


y=(1)/(8)x^2-(1)/(2)x-(1)/(2)

The directrix of the parabola is y=-3.

6.

The equation of the parabola is


y=-(1)/(32)x^2-(1)/(16)x-(289)/(32)

The directrix of the parabola is y=-1.

7.

The equation of the parabola is


y=-(1)/(12)x^2-(7)/(6)x-(109)/(12)

The directrix of the parabola is y=-2.

8.

The equation of the parabola is


y=(1)/(28)x^2-(5)/(14)x+(333)/(28)

The directrix of the parabola is y=4.

User TriumphST
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see the attached picture for the correct order:
BEFORE YOU CLICK THIS I have asked for help with this like 5 times and people have-example-1
User Ken Tan
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