Square Root Graph: Understanding x- & y- Intercepts

[lLovePhysics]

I'm confused on the graph: y=\sqrt{x}-4

I know the x- and y-intercepts which are: (0,-4) and (16,0) but I don't know if the graph extends to the negative x (3rd Quad). You can never square root a negative number and get a real number right? Therefore, does the graph just stop at the y-axis??


Btw, if you square root a negative number you can get complex numbers right? Can you not graph complex points with real points? I'm confused on this also. Just checking.

 

[Yersinia Pestis]

Remember, every real number has a positive and a negative square root. Don't feel too bad about that mistake, most graphing application programmers seem to forget as well.

So, basically, it should look a bit like y=x^2 tipped on its side.

 

[cristo]

The function sqrt(x)-4, will only have the graph in the first quadrant, since it is not a function otherwise (Look up the definition of a function if you're not convinced.) By convention, we choose the squareroot to be positive, but we could choose the negative squareroot.

 

[lLovePhysics]

The function sqrt(x)-4, will only have the graph in the first quadrant, since it is not a function otherwise (Look up the definition of a function if you're not convinced.) By convention, we choose the squareroot to be positive, but we could choose the negative squareroot.

Do you mean the first and fourth quardant? So basically the graph stops at the y-axis and doesn't go further right?

If we could choose the negative square root, why can't the graph extend further to the left (-x axis)?

 

[Teegvin]

If we could choose the negative square root, why can't the graph extend further to the left (-x axis)?

It would go back to the positive-x because it is a side-ways parabola, x=(y+4)^2.

 

[Kurdt]

I think you're confused between the choice of the range for the function and simply having a graph of f(x)=-\sqrt{x}. Choosing the negative square root is a choice of the range of the function such that we only use the part of the graph below the x axis. this will never extend to the negative x region unless we multiply it by -1.

 

[lLovePhysics]

It would go back to the positive-x because it is a side-ways parabola, x=(y+4)^2.

Ah, I see. I remember doing side ways parabolas. (y-k)^{2}=4p(x+h) right?

 

[cristo]

There seems to be some confusion here. The graph of the square root of x is not a parabola on its side-- it is half a parabola in the first quadrant of the graph. As Kurdt says, it is a restriction that we place on the range of sqrt(x) in order to ensure it is a function. Like I said earlier, anyone confused should look up the definition of a function.

 

[Yersinia Pestis]

So, how is a sideways parabola plotted? Or a circle? Is x^0.5 used? Or some other mathematical trickery?

 

[cristo]

Neither a circle nor a sideways parabola are functions.

 

[Yersinia Pestis]

So, anything that is not a function cannot be plotted? Because I've seen many graphs on Wikipedia that have more than one point of y for every x. Like elliptic curves, used in cryptography.

Now, it's almost certain that I've got completely the wrong end of the stick, but I'm not going to stop asking questions till I understand this!

 

[cristo]

So, anything that is not a function cannot be plotted? Because I've seen many graphs on Wikipedia that have more than one point of y for every x. Like elliptic curves, used in cryptography.

No, no; graphs can be plotted that are not functions. For example the graph of the unit circle can be plotted using its equation: x^2+y^2=1.

Let's return to the original topic. It is true that each real number x has two values for \sqrt{x}, namely a positive number and a negative number. However, for this to be a function we need to select one or the other of these numbers. That's why, when one sees f(x)=\sqrt{x} it really means f(x)=+\sqrt{x}-- i.e. the range is chosen to be the positive real numbers. This is a convention which makes the graph you draw a function.

 

[Yersinia Pestis]

Ah, I understand now. Thanks

And sorry to OP for threadjack.

 

[Kurdt]

Would it be useful in throwing in that functions are 1 to 1 maps. That is the functions argument produces a unique number for the entire domain of that function. As you can see for the square root case the argument produces two solutions, that is a 1 to 2 mapping. To make this a function we restrict the range of values the argument can take to make it a 1 to 1 map.

 

[lLovePhysics]

Wow this is getting confusing. Now I know that if the graph were to be a sideways parabola, then it would not be a function (vertical line test doesn't work). So you have to always draw only the top half (positive)?

Why is it you can graph a circle with both the top and bottom halves then?? So I can't draw a side ways parabola for this problem?

 

[ZioX]

Kurdt, I'm afraid I don't follow what you're saying. Functions aren't necessarily 1 to 1, they are, however, well defined.

1-1 arguments are for invertibility.

 

[Kurdt]

Kurdt, I'm afraid I don't follow what you're saying. Functions aren't necessarily 1 to 1, they are, however, well defined.

1-1 arguments are for invertibility.

Well I said it with hesitation, hence the 'Would it be useful..'. Sometimes this definition is used before more general concepts are introduced in my experience (limited as it is). Either way the OP seems not to have covered any of this stuff and thus its irrelevent.

 

[lLovePhysics]

Wow this is getting confusing. Now I know that if the graph were to be a sideways parabola, then it would not be a function (vertical line test doesn't work). So you have to always draw only the top half (positive)?

Why is it you can graph a circle with both the top and bottom halves then?? So I can't draw a side ways parabola for this problem?

 

[cristo]

Would it be useful in throwing in that functions are 1 to 1 maps. That is the functions argument produces a unique number for the entire domain of that function. As you can see for the square root case the argument produces two solutions, that is a 1 to 2 mapping. To make this a function we restrict the range of values the argument can take to make it a 1 to 1 map.

I know what you mean, but this isn't the property "one-one." For example, y=x^2 is a function-- to each value of x (i.e. each value in the domain) there corresponds only one value of y (i.e. only one value in the target). The converse (which would be required for the function to be one-one) is not satisfied.

Wow this is getting confusing. Now I know that if the graph were to be a sideways parabola, then it would not be a function (vertical line test doesn't work). So you have to always draw only the top half (positive)?

Why is it you can graph a circle with both the top and bottom halves then?? So I can't draw a side ways parabola for this problem?

You can draw a graph of anything, but it is not necessarily a function. The graph of a circle is not a function.

 

[lLovePhysics]

What's so important about functions anyways? All they have are two y outputs for every x input right? Would it make a big difference (and get marked wrong) if I drew a horizontal parabola rather than just the top half??

 

[Yersinia Pestis]

Your original question was not specified as a function, so logically you would draw the full parabola.

 

[lLovePhysics]

Your original question was not specified as a function, so logically you would draw the full parabola.

Okay, so what if a question asked me, "Is this equation a function?" Would I say no or yes?

 

[cristo]

What's so important about functions anyways? All they have are two y outputs for every x input right?

Functions have one output for each input.

Would it make a big difference (and get marked wrong) if I drew a horizontal parabola rather than just the top half??

I would mark it incorrect, yes.

Your original question was not specified as a function, so logically you would draw the full parabola.

Yes it does; when mathematicians write y=\sqrt{x} they mean the positive value of the square root of x.

 

[Yersinia Pestis]

a function looks like this:
f(x)=x^{3}+2x-7

While an ordinary equation looks like this:
y=x^{3}+2x-7

Cristo: So, how does one indicate both roots?

 

[cristo]

a function looks like this:
f(x)=x^{3}+2x-7

While an ordinary equation looks like this:
y=x^{3}+2x-7

That's just notation! They are both functions!

Cristo: So, how does one indicate both roots?

\pm\sqrt{x}

 

[Yersinia Pestis]

Gotcha, I keep forgetting about that little plussy-minusy thing.

But if, as you are saying, they are both functions, how does one represent non-functional graphs? (is my wording right? I'm trying to pick stuff up here.)

 

[Gib Z]

But wait up a second there cristo, if we had the function f(x)=\sqrt{x} then yes its only in the first quadrant. But since this function is f(x)=\sqrt{x} -4[/tex], its as if the graph of the function \sqrt{x} was moved down by 4, so it is not only in the first quadrant, but in the fourth as well. ie domain: x equal or greater than 0, range y more or equal to -4.

 

[Ariste]

Gotcha, I keep forgetting about that little plussy-minusy thing.

But if, as you are saying, they are both functions, how does one represent non-functional graphs? (is my wording right? I'm trying to pick stuff up here.)

Okay, okay. I think people are getting confused here. Here's my take on it:

A graph is simply a plot of all of the x-y pairs that satisfy a given equation. Any 2-variable equation you can think up can be graphed.

A function is a subset of graphs. It's a type of graph. A function is a graph for which every x input gives a single y output. The 'functionality,' if you will, of a graph can be tested by, as someone said, the vertical line test.

You can graph any equation, but that does not make it a function. You can graph a circle, and in order to do so you'd just draw a circle on the graph. A circle, though, is not a function.

 

[Yersinia Pestis]

So, adapting my previous example,

function:
y=\sqrt{x}
Because it always gives the positive root,

not function:
y=\pm\sqrt{x}

Because it gives two answers for each x.

Despite the curse of a useless GCSE curriculum, I think I've finally got it. Also, LaTeX is really freaking cool.

 

[cristo]

But wait up a second there cristo, if we had the function f(x)=\sqrt{x} then yes its only in the first quadrant. But since this function is f(x)=\sqrt{x} -4[/tex], its as if the graph of the function \sqrt{x} was moved down by 4, so it is not only in the first quadrant, but in the fourth as well. ie domain: x equal or greater than 0, range y more or equal to -4.

<br /> <br /> Of course. In the above I may have mixed up talking about y=sqrtx and y=sqrtx-4. However, the important point still holds-- the graph of y=sqrtx-4 is not a &quot;tipped over&quot; parabola.

 

[HallsofIvy]

Once again, no one said you can[t graph a relation that is not a function. A circle is the graph of a relation that is not a function. It fails the "vertical line test": any vertical line that passes through the graph does so only once.
The graph of (y+4)²= x is a "parabola lying on its side" but y is NOT a function of x. If x= 4, then y can be either -2 or -6: the vertical line x= 4 passes through the graph at both (4,-2) and (4,-6).

However, the function y= \sqrt{x} is defined as "the positive number whose square is x" and is a function. Then y= \sqrt{x}- 4 is the part of a parabola that starts at (0,-4) and rises upward and two the right. The functions domain is [0,\infty) and its range is [-4,\infty).

 

[soadgirl678]

I recently got marked wrong on a calc test for this... my teacher gave us y= √ x (not ±) and x=3-2ysquared. We were supposed to find the area between the two curves and I got the wrong answer since y= √ x (not ±) was only the positive half of the parabola. She argued that it was the full parabola since we should've turned it into x= ysquared. Is that right? can you just suare both sides and magically get both halfs of the parabola? This doesn't make sense to me.

 

[aznHypnotix]

The square root of x on a graph starts from zero and is infinite. X > zero for the square root of x where x is all real numbers. Square root of x - 4 is x > -4 where x is all real numbers and is infinite.

 

[kmikias]

Remember, every real number has a positive and a negative square root. Don't feel too bad about that mistake, most graphing application programmers seem to forget as well.

So, basically, it should look a bit like y=x^2 tipped on its side.

What do you mean bit like y=x^2.
you mean half parabola in 4th quadrant.

 

[NoMoreExams]

Think of it as the branch of y = x^2 that is in the 1st quadrant "flipped" over y = x.

 

[HallsofIvy]

What's so important about functions anyways? All they have are two y outputs for every x input right? Would it make a big difference (and get marked wrong) if I drew a horizontal parabola rather than just the top half??

No, they have one y output for every x input! If you were specifically asked to graph the function y= \sqrt{x} and graphed the entire parabola, yes you would be marked wrong. If you were asked to graph x= y² (so y is a "relation", not a function of x, though now x is a function of y) then you should graph the entire parabola.

There is nothing terribly important about "functions" (except that they are somewhat simpler than "relations") in mathematics but they tend to be very important in applications of mathematics to science because of the requirement of "repeatability": if you do an experiment twice, with everything set up exactly the same way, you should get exactly the same result- one input, one output. "This causes that" gives functions.

Think of it this way: If you were to go to a store and find different products that had the same price, you would not be surprised, right? The "product" is not a function of the price. On the other hand, if you found exactly the same product, same size, brand, and everything, in the same store for two different prices, you would know that something was wrong: the price is a function of the product.