Probably not. Meaning that parabolas can show up on the GRE, but probably won’t show up on the GRE you are taking. That is my initial take based on forums and students experiences. Still, parabolas, along with absolute value graphs, are included in the Practicing to Take the Revised GRE book. ETS obviously didn’t put them there to kill more trees. So, like compound interest and the quadratic equation, parabolas may show up.

What this means for your study plans is that you should only worry about parabolas if you are looking to score above 90%, and have already brushed up on the other concepts on the new GRE math. Meaning, if you are still struggling with number properties or circles, focus there.

That said, below are some important things you have to know about parabolas.

## Introduction to parabolas on the GRE

## Symmetry

If you draw a line dividing a parabola in the middle, the parabola will be split into two equal halves. This line is known as the axis of symmetry. The shape of the parabola to the right of the axis symmetry is identical to the shape of the parabola the left of it.

The axis of symmetry will either be a vertical line or a horizontal line, but the effect will always be the same: to split the parabola into two equal halves.

## The equation of a parabola

The equation for a parabola will always contain a coefficient, meaning that x is always squared: . This may not be too helpful, so just think of it this way: , when graphed, is a parabola. On the other hand, when the x is not squared, say,

The equation for a parabola can also be written as

This form is also known as the vertex form and is expressed as

So how does all of this pertain to the new GRE? Well, remember the line of symmetry? To find it, we simply look at the value of h. In the example above, we have

Finally, there is the vertex, which is either the highest or lowest point depending on the equation (see horizontal vs. vertical, upwards vs. downwards). The vertex can always be represented by

Notice that the k is in the place of the 1 on the equation to the left. Therefore

## Intersecting lines

The point at which a straight line intersects a parabola can be found by setting the equation for the line and the equation for the parabola equal to each other. If a line intersects a parabola set the two lines equal to each other.

For instance the line,

To find y, we simply plug

Sometimes a question will simply ask you at how many points a line intersects a parabola. A line can intersect a parabola at zero points, one point, or two points.

Sometimes simply graphing out the parabola and the line is the easiest way to answer such a question.

## Types of parabolas: horizontal vs. vertical, upwards vs. downwards

Let’s have a look at two different parabolas:

The two equations are identical, save that the x and y have been swapped. Whenever a parabola is equal to x it is horizontal, meaning that the axis of symmetry is horizontal.

When the parabola is equal to y, the axis of symmetry is vertical.

In the case of

Now have a look at the following equations:

This pair of equations is identical to the ones above, except for there is a negative in front of both

If this all seems very abstract take a look at the video above.

## Skinny vs. fat parabolas

Finally, a parabola can change its shape depending on whether the coefficient (or number) next to

(Skinny and fat are my terms not standard math-ese. So if you whip them out in front of math inclined folk you are likely to get some very quizzical looks).

## Takeaway

There is much to learn on parabolas, as you can see from the post above. However, there is little likelihood that a parabola will even show up on the test so study only if you are aiming for a top math score.

The accompanying video should hopefully make what I covered in this post far less abstract.

### Most Popular Resources

Is it possible for a parabola to have a line of symmetry as y=x or y = -x ? if yes then how can we find other intercepts if one is given?

Hi Himan,

As Chris mentions in this blog post, the axis of symmetry will always be either a horizontal or vertical line, so you won’t have an axis of symmetry of y=x or y=-x. For more information on this and how to find the intercepts, I recommend that you check out the Khan Academy videos that cover this subject: https://www.khanacademy.org/math/algebra/quadratics

Great article – inclusion of diagrams would add an extra layer of icing to the well done cake!

I’m glad you like the article. And I’ll check with our content team to see if we can add in some visuals. 🙂

Hi Chris

I have just completed the Graphs of quadratics lesson from the math lessons on Magoosh. In that it was told that for a given vertex of a parabola the line of symmetry is always the vertical line passing through the vertex which is x= (x co ordinate of the vertex) however I doubt that, it may be possible to have a slant line passing through the vertex to be the line of symmetry for the parabola. Can you please clarify this?

Hi Ujwal. Good question! The line of symmetry for a parabola will always be a straight vertical line. The vertical line divides the parabola into two congruent or halves, where for every value of y, there are two values of x: x and -x. 🙂

Hi Chris,

Can you please update the video for parabola? I am not able to find it and my test is due in the coming month. So, I want to be accoutered with all the mathematical concepts.

Thanks in advance.

Shweta

Hope this helps 🙂

https://www.youtube.com/watch?v=gyGPc3UI_5M

Here is the Video Chris is mentioning which seem to have disappeared

Youtube : https://www.youtube.com/watch?v=gyGPc3UI_5M

Titled as “GRE Math: Parabolas ” incase you want to search on google.

Thanks Puneet for finding that 🙂

Hi Chris,

I was one of the those people who ended up with a parabola on her test, and I was absolutely flummoxed (that semester of pre-calc four years ago clearly did not stick). I’m now terrified there is going to be another one on my retake. The question I had asked me to choose from a group of five graphs the shape of the parabola that matched the equation given. Do the tips you outlined above apply to a question like that?

Thanks!

Hi Liv,

Sorry about that! I think the takeaway is the GRE isn’t trying to make the parabolas too difficult–they just want you to have a general sense of how things work. The question you cited is actually more straightforward than the info. above. Though the info. above should help you answer such a question. However, I didn’t really touch on the physical properties of the parabola so translating from an equation to a coordinate plane still takes some practice.

Let’s look at the following equations:

A. x^2 – 1

B. 1/4x^2 + 3

C. 4x^2

The number next to the x^2 in each equation (known as the coefficient) determines the width/narrowness of a parabola. The lower the number the “fatter” the parabola. So a coefficient of 1/4 will yield a much wider parabola than 4, which will make a “skinny” parabola. Therefore, in the equations above, (A) is of average girth, (B) is wide, and (C) is narrow.

The number that is by itself (in (A) that number would be -1) is where the parabola crosses through the y-axis. Therefore, (A) would have the bottom point of the parabola at -1, (B) at positive 3, and (C) would have the vertex on the origin.

Finally, if there is a negative next to the coefficient and the x^2, then the parabola will be upside-down. (A), (B) and (C) are all right side up.

I’d encourage you just to graph at some of these equations. Play around with the equations, add some negatives, change some coefficients, and you should quickly get the hang of it.

Hope that helps!

Chris,

Will I need to worry about converting a quadratic equation from general form (ax^2+bx+c) to vertex form ((x-h)^2+k)? If so don’t you do this by completing the square?

I don’t quite recall all the steps involved with completing the square.

Hi Nicholas,

I’ve never seen a parabola question on the GRE that requires you to do so (and that’s mainly because there just aren’t that many parabola questions out there). So it doesn’t hurt to learn how to complete the square. And it’s not too difficult.

Say I have x^2 + 6x + 13. To get this into parabola form, divide the ‘b’ (which is 6) by 2 and then square. This will give you 9. Thus you can write the original equations as the following:

x^2 + 6x + 9 + 4 —> (x + 3)^2 + 4

Hope that helps!

For those who are still lingering around:

X co-ordinate of the vertex= -b/a

(No idea if it’ll help in the actual exam, I’ve been scared of “completing the square” and factorisation since the eight grade)

I would also like to know where I can find the mentioned video. 🙂

Hi Chris,

Maybe I missed something, but you mention the following: “If this all seems very abstract take a look at the video above.” Which video are you referring to and where can I find it? I’m pretty weak with parabolas so a video explaining the concepts above would be helpful! 🙂 Thank you!

The video seems to have disappeared :(. I will have to rerecord a new one — which might take a little time :). The thing is I wouldn’t worry about parabolas too much. There is less than a 50% chance you will even see one parabola-related question during the entire test.

Hope that helps allay your worries a little :)!