Will there be Parabolas on the GRE?

January 26, 2012

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, x-1=2 , we have a simple line.

The equation for a parabola can also be written as . (This would turn out to be if written in the format in the preceding paragraph).

This form is also known as the vertex form and is expressed as . The reason this is call vertex form is it makes finding the vertex very easy.

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 , therefore h is equal to 1.

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 (h, k) . Since h is already 1, we need to find k. This becomes straightforward if we put the following two equations next to each other:

and

Notice that the k is in the place of the 1 on the equation to the left. Therefore k = 1 . which is also 1. Hence the vertex.

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, y=2x-1 and the parabola y=x^2 , both equal y so we can set them to equal to each other: . We get , which equals (x-1)^2 . Therefore x=1 .

To find y, we simply plug x=1 into either equation (they will always be the same). The answer then is 1. So the line and the parabola intersect at (1,1) .

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:

x=y^2-1
y=x^2-1

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 x=y^2-1 the parabola opens to the right. In the case of y=x^2-1 the parabola opens upward.

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 y^2 and x^2 . This negative flips the direction of each parabola.

opens to the left and opens downwards.

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 x^2 is less than or greater than one. For instance, in the equation
y=3x^2 the coefficient is 3, so we will have a ‘skinny’ parabola—that is, the great the coefficient the skinnier the parabola. On the other hand, in $y={1/3}x^2$ , the coefficient is 1/3 so we will have a ‘fat’ parabola.

(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.

Author

Chris Lele

Chris Lele is the Principal Curriculum Manager (and vocabulary wizard) at Magoosh. Chris graduated from UCLA with a BA in Psychology and has 20 years of experience in the test prep industry. He’s been quoted as a subject expert in many publications, including US News, GMAC, and Business Because. In his time at Magoosh, Chris has taught countless students how to tackle the GRE, GMAT, SAT, ACT, MCAT (CARS), and LSAT exams with confidence. Some of his students have even gone on to get near-perfect scores. You can find Chris on YouTube, LinkedIn, Twitter and Facebook!

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19 responses to “Will there be Parabolas on the GRE?”

Himan

July 1, 2017

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?

Reply
1. Magoosh Test Prep Expert
  
  July 10, 2017
  
  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
  
  Reply
Prashant

April 19, 2017

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

Reply
1. Magoosh Test Prep Expert
  
  April 21, 2017
  
  I’m glad you like the article. And I’ll check with our content team to see if we can add in some visuals. 🙂
  
  Reply
Ujwal

June 19, 2016

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?

Reply
1. Magoosh Test Prep Expert
  
  June 19, 2016
  
  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. 🙂
  
  Reply
Shweta

September 24, 2014

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

Reply
1. Chris Lele
  
  September 24, 2014
  
  Hope this helps 🙂
  
  https://www.youtube.com/watch?v=gyGPc3UI_5M
  
  Reply
Puneet

August 14, 2014

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.

Reply
1. Chris Lele
  
  August 14, 2014
  
  Thanks Puneet for finding that 🙂
  
  Reply
Liv

August 27, 2013

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!

Reply
1. Chris Lele
  
  August 28, 2013
  
  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!
  
  Reply
Nicholas Thomas

August 19, 2013

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.

Reply
1. Chris Lele
  
  August 20, 2013
  
  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!
  
  Reply
2. Asimio
  
  October 26, 2015
  
  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)
  
  Reply
  1. Mohan Giri
    
    May 31, 2019
    
    x co-ordinate of the vertex = -b/2a I guess
    
    Reply
Manuel

July 28, 2013

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

Reply
Inny

July 24, 2013

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!

Reply
1. Chris Lele
  
  July 30, 2013
  
  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 :)!
  
  Reply