Parabolas - Pre-Calculus

The first step of the problem is to find the axis of symmetry using the following formula:

Where a and b are determined from the format for the equation of a parabola:

We can see from the equation given in the problem that a=1 and b=-3, so we can plug these values into the formula to find the axis of symmetry of our parabola:

Keep in mind that the vertex of the parabola lies directly on the axis of symmetry. That is, the x-coordinate of the axis of symmetry will be the same as that of the vertex of the parabola. Now that we know the vertex is at the same x-coordinate as the axis of symmetry, we can simply plug this value into our function to find the y-coordinate of the vertex:

So the vertex occurs at the point:

Find the axis of symmetry and the vertex of the parabola given by the following equation:

To find the axis of symmetry of a parabola in standard form, , use the following equation:

So...

This means that we have an axis of symmetry at . Or, to put it more plainly, at we could draw a vertical line which would perfectly cut our parabola in half!

So, we are halfway there, now we need the coordinates of our vertex. We already know the x-coordinate, which is 7. To find the y-coordinate, simply plug 7 into the parabola's formula and solve!

This makes our vertex the point

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 .

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.

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

In order to determine which direction the parabola opens, we must first put the equation in standard form, which can be expressed in one of the following two ways:

If the equation is for as in the first above, the parabola opens up if is positive and down if is negative. If the equation is for as in the second above, the parabola opens right if is positive and left if is negative. Rearranging our equation, we get:

We can see that our equation is for , which means the parabola will open either left or right. The sign of the first term is negative, so this parabola will open to the left.

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

We can see that our equation has all of the components of the first form above, so now all we must do is some algebra to rearrange the equation and express the function y in terms of x. We start by simplifying the fraction on the left side of the equation, and then we isolate y to give us the equation of the parabola in standard form:

For the function

The parabola opens upwards if a>0

and downards for a<0

Because

The parabola opens upwards.

For the function

The parabola opens upwards if a>0

and downards for a<0

Because

The parabola opens downwards.

The standard form for a parabola is in the form:

The coefficient of the term determines whether if the parabola opens upward or downward. Since the term in the function is , the parabola will open downward.

Use the FOIL method to determine in its standard form for parabolas, which is .

Regroup the terms.

Since the coefficient of the term is negative, the parabola will open downward.

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 .

First, write the equation of the parabola in standard form.

Determine the values of the coefficients. The value of the y-intercept is 4, which means that .

Write the vertex formula.

The given point of the vertex is , which indicates that:

Substitute the value of the point and into the standard form.

Substitute this value into to determine the value of .

Substitute the values of coefficients into the standard form of the parabola.

The standard form of a parabola is .

Factorize the right side of , and simplify.

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 .

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 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

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 .

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 into the standard form.

Isolate the terms on one side.

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

Factor both sides of the equation to get the standard equation for the parabola.

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

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.

For the given parabola, the vertex is and . This means the parabola is opening to the left and that the directerix will be located unit to the right of the vertex. The directerix is then .