## How to draw y^2 = x – 2?

[09 Apr 2009]

Nuaja, a subscriber to the IntMath Newsletter, wrote recently:

How do I know how the graph should look like: For example: y2 = x – 2?

The first thing I recognize in that equation is the y2 term, which tells me it will be a parabola. (It won’t be a circle, ellipse or hyperbola because there is an x term, but no x2 term. See Conic Sections.)

Let’s start with the most basic parabola y = x2 and build up to the required answer.

### Example 1: y = x2

You could draw up a table and calculate the y-values for a set of x-values, like this:

 x -4 -3 -2 -1 0 1 2 3 4 y 16 9 4 1 0 1 4 9 16

This gives us a series of points (-4,16), (-3,9), (-2,4) up to (4,16).

You then join these dots with a smooth curve and get something like the following.

Notice that the vertex of the parabola (the “pointy” end) is at the origin, (0, 0).

Now for all the curves that I draw below, I’m not going to draw up a table. It becomes tedious, and it can lead to incorrect graphs. It is better to be able to recognize the graph type (from the equation) and then know how to sketch it in the right place and with the right orientation.

I will consider the effect of small changes to the equation and then sketch my curve.

All of the following graphs have the same size and shape as the above curve. I am just moving that curve around to show you how it works.

### Example 2: y = x2 − 2

The only difference with the first graph that I drew (y = x2) and this one (y = x2 − 2) is the “minus 2″. The “minus 2″ means that all the y-values for the graph need to be moved down by 2 units.

So we just take our first curve and move it down 2 units. Our new curve’s vertex is at −2 on the y-axis.

Next, we see how to move the curve up (rather than down).

### Example 3: y = x2 + 3

The “plus 3″ means we need to add 3 to all the y-values that we got for the basic curve y = x2. The resulting curve is 3 units higher than y = x2. Note that the vertex of the curve is at (0, 3) on the y-axis.

Next we see how to move a curve left and right.

### Example 4: y = (x − 1)2

Note the brackets in this example – they make a big difference!

If we think about y = (x − 1)2 for a while, we realize the y-value will always be positive, except at x = 1 (where y will equal 0).

Before sketching, I will check another (easy) point to make sure I have the curve in the right place. Putting x = 0 is usually easy, so I substitute and get

y = (0 − 1)2

= 1

So the curve passes through (0, 1).

Here is the graph of y = (x − 1)2.

### Example 5: y = (x + 2)2

With similar reasoning to the last example, I know that my curve is going to be completely above the x-axis, except at x = −2.

The “plus 2″ in brackets has the effect of moving our parabola 2 units to the left.

### Rotating the Parabola

The original question from Anuja asked how to draw y2 = x − 4.

In this case, we don’t have a simple y with an x2 term like all of the above examples. Now we have a situation where the parabola is rotated.

Let’s go through the steps, starting with a basic rotated parabola.

### Example 6: y2 = x

The curve y2 = x represents a parabola rotated 90° to the right.

We actually have 2 functions,

y = √x (the top half of the parabola); and

y = −√x (the bottom half of the parabola)

Here is the curve y2 = x. It passes through (0, 0) and also (4,2) and (4,−2).

[Notice that we get 2 values of y for each value of x larger than 0. This is not a function, it is called a relation.]

### Example 7: (y + 1)2 = x

If we think about the equation (y + 1)2 = x for a while, we can see that x will be positive for all values of y (since any value squared will be positive) except y = −1 (at which point x = 0).

In the equation (y + 1)2 = x, the “plus 1″ in brackets has the effect of moving our rotated parabola down one unit.

### Example 8: (y − 3)2 = x

Using similar reasoning to the above example, the “minus 3″ in brackets has the effect of moving the rotated parabola up 3 units.

Finally we are ready to answer the question posed by Nuaja.

### Example 9: y2 = x − 2

You can hopefully imagine what is going to happen now. We have a y2 term , so it means it will be a rotated parabola.

When x = 2, y = 0. The value of x cannot be less than 2, otherwise when we try to evaluate y we would be trying to find the square root of a negative number. Since out numbers are all real numbers, x must be greater than or equal to 2.

The “minus 2″ term has the effect of shifting our parabola 2 units to the right.

I hope you can see now that if the equation was y2 = x + 2 (with a “plus”), then we would need to shift our rotated parabola to the left by 2 units.

So Nuaja, I hope that answers your question.

Like all things, the best way how to learn graph sketching is through practice. Also, be observant and note the effect of plus, minus and brackets in each example.

### 48 Comments on “How to draw y^2 = x – 2?”

1. hassan says:

thanks, nice tip…

2. Nandan says:

Remarkable

3. MAVIS says:

I hv learned a lot, thanks. The problem with me is integration.

4. amane says:

thank you for your assistance.

5. kiran says:

thanks for help in maths nd i hav learned alot from this and my problem is planes

6. max says:

One word, FANTASTIC.

looking forward to the next e-mail.

7. hosea says:

Thanks for your explanation, it has assisted me much. Keep it up.

8. Munzir says:

thanks for the tip i nvr knew that sketching graphs are so easy and not so tedious. thank you very much

9. ricar says:

TNX A LOT! KEEP UP..

10. Masereka Hosea says:

11. Murray says:

Hi Masereka. Do you mean “Laplace”? You can find a chapter on it here: Laplace Transformation.

12. chirag hira says:

i do not understand how you drew the graph for y = x2( example 1)

can you please give me some pointers on how to go about doing these things

i am currently in grade 10 (std8)

plaese if any one can help

thanks alot

13. Murray says:

Hi Chirag. The table for example 1 is obtained by substituting the different x-values into y = x2. So for example, when x = -2, then y = (-2)2 = 4. This is graphed by putting a dot on the point (-2,4).

We then put dots for all the other points in the table (like (-4,16), then (-3,9) and so on. We join the dots with a smooth curve and end up with the parabola shown.

Hope that helps.

14. chirag hira says:

thanks alot zac that realy did help and made me think

thanks for all the help
chirag

15. Zach says:

I was wondering how to graph y^2=x in a graphing calculator

Thanks, Zach

16. Murray says:

Hi Zach

Since y^2 = x − 2 is a relation (has more than 1 y-value for each x-value) and not a function (which has a maximum of 1 y-value for each x-value), we need to split it into 2 separate functions and graph them together. So the first one will be y1 = √(x − 2) and the second one is y2 = −√(x − 2). When you graph these on the same axis, it will give the required result.

17. nimitz says:

thanks i know it

18. nandan says:

very good

19. Mike Theodore says:

Great examples.

20. peter says:

thanks so much this is so helpful!!

21. mohamed says:

it’s easy to understand, sketching is so important so we need programm can draw the function in moments in this good and best website.

22. deedee says:

hey, thanks alot 4 d examples! it did help….
but i’m still confused on how d x and y values were derived from just an equation.
i hope u can pls shed more light on that…
thanks

23. Murray says:

@deedee: Are you referring to the table of values in Example 1?

They just come from substituting in values. The first one is x = -4, and substituting that into y = x^2 gives y = (-4)^2 = 16.

24. Shamsa Aden says:

Hi,
I woul like to know how to find the simplest general formua of parabolic graph? and you can find the tangent line in it?

thank you.

25. Murray says:

Hi Shamsa. The simplest general formula for a parabola is given in the article: y = ax2 + bx + c.

You can find how to differentiate polynomials here: Derivatives of Polynomials.

Later on that page is an example of a tangent line.

26. jr says:

how do u find the parabola if all the numbers are positive?????

27. Murray says:

Hello jr. Do you mean if the question was y^2 = x + 2? Well, instead of moving the parabola to the right by 2 units (like I did at the end in the above example), you would move it to the left by 2 units, as mentioned below the final graph.

28. Caitlyn says:

How would you draw a graph if the question is only half a parabola?
y^2= √(-8x)

29. Murray says:

It’s actually a “full” parabola, Caitlyn. The negative in front of the x just flips the parabola through the y-axis so the arms face to the left, instead of the right like the final examples above.

30. PdfoK FilE says:

Gret site. Thanks. I agree that this site is rather cognitive and useful especially for professionals.

31. Jovan says:

But how to draw a y=2-x^2..

32. Murray says:

Hi Jovan

Well, first can you sketch y = -x^2?

Then, what does the 2 out the front do?

33. tamirat tesfaye says:

How i can draw the graph of parabola when only two points are given.

34. Murray says:

@Tamarat: There are an infinite number of parabolas passing through 2 points. You always need 3 points to determine a parabola, and you also have to specify the direction of the axis (vertical, horizontal or at an angle).

35. gagangc says:

sir,
if i want to plot same function by finding intercepts and slopes,then y intercept will involve complex numbers. what it means?,cant i plot it if i take my y axis as imaginary axis?,in that case graph would be different.

36. Murray says:

@Gagangc: Interesting question! All the plots on this page assume real axes. You could, I guess, plot the ones involving complex numbers on the 3D complex plane, and the result would be something like the ones on this page: exponential 3D plots.

37. gagangc says:

Sir,

i tried sketching xy = 4 (hyperbola),using curve sketching by differentiation,i got x and y intercepts as infinity which i understood.but problem came when i tried finding maxima and minima as i had to equate (-4)/(x^2) to 0.what is it telling to me??i could not come to any conclusion.problem got bigger when i found second derivative.help me find solution.

38. Murray says:

@gagangc: A hyperbola in this orientation doesn’t have a maximum or minimum! You’ll find an example similar to your question on this page: Hyperbola.

39. gagangc says:

Sir,
If the real roots of a quadratic eqn are the x-intercepts, what do imaginary roots represent graphically?.(as in x^2+9).i am bit confused regarding this,help me to get out of this problem.

40. Murray says:

@gagangc: You can’t represent complex roots on a real cartesian plane. However, we can plot complex numbers on the complex plane (with “reals” on the horizontal axis and “imaginary” on the vertical axis).

41. Brenda Radnedge says:

Finding these graph difficult to do. Could someone help me with sketch the graph of y squared = 25 – x squared I need to know how to work it out so we can apply it to other values. thanks.

42. Murray says:

Hi Brenda

Just add x^2 to both sides and then check out this:
http://www.intmath.com/plane-analytic-geometry/3-circle.php

43. Cheryl says:

how to sketch the graph of x=(y-2)^2 +4 ?? thank you

44. Murray says:

Hello Cheryl. The answer for your question is contained in the article!

Is your graph a parabola?
What orientation does it have? (Legs up? Right? Left? Down?)
What does the “-2″ in the brackets do?
What does the “+4″ do?

45. arthur says:

y=(x-2)*2