Question

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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.

Answer 1

see the attached picture for the correct order:

Answer 2

Answer:

The order of equation is

$y=\frac{1}{8}x^2-\frac{1}{2}x-\frac{1}{2}$

$y=-\frac{1}{12}x^2-\frac{7}{6}x-\frac{109}{12}$

$y=-\frac{1}{32}x^2-\frac{1}{16}x-\frac{289}{32}$

$y=\frac{1}{20}x^2+\frac{2}{5}x+\frac{29}{5}$

$y=-\frac{1}{16}x^2+\frac{3}{4}x-\frac{21}{4}$

$y=\frac{1}{8}x^2+\frac{3}{4}x+\frac{41}{8}$

$y=-\frac{1}{52}x^2-\frac{4}{13}x-\frac{146}{13}$

$y=\frac{1}{28}x^2-\frac{5}{14}x+\frac{333}{28}$

Step-by-step explanation:

If a parabola is defined as

$y=ax^2+bx+c$

then the directrix of the parabola is

$y=-\frac{b^2}{4a}-\frac{1}{4a}+c$

1.

The equation of the parabola is

$y=\frac{1}{8}x^2+\frac{3}{4}x+\frac{41}{8}$

The directrix of the parabola is

$y=-\frac{(\frac{3}{4})^2}{4(\frac{1}{8})}-\frac{1}{4(\frac{1}{8})}+(\frac{41}{8})$

$y=2$

Similarly find the directrix of each parabola.

2.

The equation of the parabola is

$y=\frac{1}{20}x^2+\frac{2}{5}x+\frac{29}{5}$

The directrix of the parabola is $y=0$

3.

The equation of the parabola is

$y=-\frac{1}{52}x^2-\frac{4}{13}x-\frac{146}{13}$

The directrix of the parabola is $y=3$

4.

The equation of the parabola is

$y=-\frac{1}{16}x^2+\frac{3}{4}x-\frac{21}{4}$

The directrix of the parabola is $y=1$

5.

The equation of the parabola is

$y=\frac{1}{8}x^2-\frac{1}{2}x-\frac{1}{2}$

The directrix of the parabola is y=-3.

6.

The equation of the parabola is

$y=-\frac{1}{32}x^2-\frac{1}{16}x-\frac{289}{32}$

The directrix of the parabola is y=-1.

7.

The equation of the parabola is

$y=-\frac{1}{12}x^2-\frac{7}{6}x-\frac{109}{12}$

The directrix of the parabola is y=-2.

8.

The equation of the parabola is

$y=\frac{1}{28}x^2-\frac{5}{14}x+\frac{333}{28}$

The directrix of the parabola is y=4.