## How Do I Know Which Direction the Parabola Opens?

#### From the Value of a

There are a couple of ways we can determine the direction the parabola opens. The a-value is one way. If a > 0, then the parabola opens upward. If a < 0, then the parabola opens downward. Recall that when the parabola is given in standard form, y = ax² + bx + c, the a-value is the leading coefficient.

Example 1: If y = 3x² -4x + 1, determine the parabola’s orientation.

Solution: In this example, a = 3, and because 3 > 0, the parabola opens upward. ■

The parabola can also be given in vertex form. Recall the vertex form of a parabola is given as y = a(x-h)² + k. We still use the a-value and use the same test when the equation of the parabola is given to us in vertex form.

Example 2: If y = -2(x+3)² – 3, determine the parabola’s orientation.

Solution: In this case, a = – 2 < 0. So this parabola opens downward.   ■

#### From the Relative Position of the Directrix and Focus

Another way we can determine a parabola’s orientation is by knowing the relative vertical position between the focus and directrix. When the focus is above the directrix, then the parabola opens upward. When the focus is below the directrix, the parabola opens downward.

Example: A parabola’s focus is F(1,4) and directrix is y=7. Determine the parabola’s orientation.

Solution: Because the directrix is y = 7 and the y-coordinate of the focus is y = 4, the directrix is above the focus. Thus, the parabola opens downward.   ■

#### From the Second Derivative and the Concavity Test

For you fancy-pants calculus students, you do not need to know anything about parabolas in general to know what direction it opens. Given the equation of the parabola, in either standard form or vertex form, you can perform a test for concavity by taking the second derivative of the given function. On a quadratic polynomial, the second derivative always yields a constant. When it yields a positive constant, we know the original function is everywhere concave up. When it yields a negative constant, the original function is everywhere concave down. In other words, when the second derivative of the equation of our parabola is positive, the parabola opens upward; when it is negative the parabola opens downward. That’s all there is to it. Here’s how it works in practice.

Example 1: If y = 2x² + 4x – 1, determine the parabola’s orientation.

Solution: Because y = 2x² + 4x – 1, then y’ = 4x + 4, and y”= 4. Because y” = 4 > 0, the parabola is everywhere concave up. Thus, the parabola opens upward.   ■

Here is an example of determining the parabola’s orientation with the concavity test when the original equation is given in vertex form.

Example 2: If y = -2(x-1)² + 3, determine the parabola’s orientation.

Solution: If y = -2(x-1)² + 3, then y’ = -4(x-1), and y” = -4. Because y” = -4 < 0, the parabola is everywhere concave down. So the parabola opens downward.   ■

Here’s what we learn when we apply the concavity test to the general quadratic polynomial.

Example 3: If y = ax² + bx + c, for a, b, and c real numbers, determine the parabola’s orientation.

Solution: Because y = ax² + bx + c, y’ = 2ax + b, and y”= 2a. Note that y” > 0 if a > 0, and y” < 0 if a < 0. So the value of a determines whether the parabola is everywhere concave up or everywhere concave down. Thus, the parabola opens upward whenever a > 0. The parabola opens downward whenever a < 0.   ■

• http://www.protutorcompany.com/ The ProTutor Company

This is a great question we received last night. We tried to send a reply, but are having trouble reaching Yahoo! users. Here’s the question she had:

“A parabola is defined by the equation -2y² = 4x. In which direction will the parabola open?”

There are a couple of ways to determine a parabola’s “orientation” (i.e. the direction the parabola opens). Possibly the easiest way for the type of problem you have:

-2y² = 4x,

is to zero in on the fact that you have a y² floating around, instead of an x². Usually with the “vertically oriented” parabolas (that open upward or downward), we have something like y = ax², and the “2” exponent is attached to the x. In those cases, it’s an x² that is floating around.

In the case when we have x = ay², we have a y² floating around. When we do, we have a parabola that is “horizontally oriented” — i.e., it is on its side, and we know this parabola opens to the left or to the right.

Furthermore, when we have a horizontal parabola in the form x=ay², we know that if a > 0, the parabola opens to the right. If a < 0, the parabola opens to the left.

If this rule is unfamiliar to you, try plotting a parabola like x = 3y² by using a table of values. Then do the same for x=-3y². You will be a believer in no time!

In your problem, we have -2y² = 4x, or if we divide both sides by 4, to get x by itself, we get x = (-1/2) y². In this case, we have a parabola on its side and our "a" value is (-1/2), which is negative. So we know our parabola opens to the left.

Again, we can check this by constructing a table of values. For example, if we let y = 0, 2, 4, -2, -4, we find the corresponding x-values to be 0, -2, -8, -2, -8, respectively. If we plot those five points: (0,0), (-2,2), (-8,4), (-2,-2), (-8,-4), you will indeed see that we have a parabola that opens to the left!