The answer is because it is the only degree-2 polynomial with a negative leading coefficient.

Recall the standard equation of a horizontal parabola:

, where is the vertex and is the focal length.

Start by isolating the terms.

Complete the square on the left. Make sure to add the same amount to both sides of the equation!

Factor both sides of the equation to get the standard form of a horizontal parabola.

Recall the standard form of the equation of a vertical parabola:

, where is the vertex of the parabola and gives the focal length.

When , the parabola will open up.

When , the parabola will open down.

For the parabola in question, the vertex is and . This parabola will open up. Because the parabola will open up, the directerix will be located units down from the vertex. The equation for the directerix is then .

To be in standard form, the equation for a parabola must be written in one of the following ways:

OR

THe problem given has the square around the x term, so it's going to end up loking like the standard form on the left.

First, we square the right side

Lastly, we need the y by itself, so we add 3 to both sides

Recall the standard form of the equation of a vertical parabola:

, where is the vertex of the parabola and gives the focal length.

When , the parabola will open up.

When , the parabola will open down.

For the parabola in question, the vertex is and . This parabola will open up. Because the parabola will open up, the directerix will be located unit down from the vertex. The equation for the directerix is then .

In order to determine which way this parabola, group the variables in one side of the equation. Add on both sides of the equation to isolate .

Because the equation is in terms of , the parabola will either open left or right. Notice that the coefficient of the term is negative.

The parabola will open to the left.

To find the focus from the equation of a parabola, first set the equation to resemble the form where represents any numerical value.

For our problem, it is already in this form.

Therefore,

.

Solve for then .

The focus for this parabola is given by .

So, is the focus of the parabola.

The directrix is represented as .

Therefore, the directrix for this problem is .

Recall the standard form of the equation of a vertical parabola:

, where is the vertex of the parabola and gives the focal length.

When , the parabola will open up.

When , the parabola will open down.

Start by putting the equation in th estandard form of the equation of a vertical parabola.

Isolate the terms to one side.

Complete the square for the terms. Remember to add the same amount on both sides!

Factor out both sides of the equation to get the standard form of a vertical parabola.

For the parabola in question, the vertex is and . This parabola will open down. Because the parabola will open down, the directerix will be located units above the vertex. The equation for the directerix is then .

Recall the standard form of the equation of a horizontal parabola:

, where is the vertex of the parabola and is the focal length.

When , the parabola opens to the right.

When , the parabola opens to the left.

Start by putting the equation into the standard form of the equation of a horizontal parabola.

Isolate the terms on one side.

Complete the square. Remember to add the same amount to both sides of the equation!

Factor both sides of the equation to get the equation in the standard form.

For the given parabola, the vertex is and . This means the parabola is opening to the right and that the focus will be located units to the right of the vertex. The focus is then located at .

Recall the standard form of the equation of a vertical parabola:

, where is the vertex of the parabola and gives the focal length.

When , the parabola will open up.

When , the parabola will open down.

For the parabola in question, the vertex is and . This parabola will open down. Because the parabola will open down, the focus will be located units down from the vertex. The focus is then located at