Square root graph

  • #26
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.)
 
  • #27
Gib Z
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But wait up a second there cristo, if we had the function [itex]f(x)=\sqrt{x}[/itex] then yes its only in the first quadrant. But since this function is [itex]f(x)=\sqrt{x} -4[/tex], its as if the graph of the function [itex]\sqrt{x}[/itex] 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.
 
  • #28
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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.
 
  • #29
So, adapting my previous example,

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

not function:
[tex]y=\pm\sqrt{x}[/tex]

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.
 
  • #30
cristo
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But wait up a second there cristo, if we had the function [itex]f(x)=\sqrt{x}[/itex] then yes its only in the first quadrant. But since this function is [itex]f(x)=\sqrt{x} -4[/tex], its as if the graph of the function [itex]\sqrt{x}[/itex] 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.

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 "tipped over" parabola.
 
  • #31
HallsofIvy
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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 [itex]y= \sqrt{x}[/itex] is defined as "the positive number whose square is x" and is a function. Then [itex]y= \sqrt{x}- 4[/itex] is the part of a parabola that starts at (0,-4) and rises upward and two the right. The functions domain is [itex][0,\infty)[/itex] and its range is [itex] [-4,\infty)[/itex].
 
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  • #32
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.
 
  • #33
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.
 
  • #34
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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.
 
  • #35
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Think of it as the branch of y = x^2 that is in the 1st quadrant "flipped" over y = x.
 
  • #36
HallsofIvy
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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 [itex]y= \sqrt{x}[/itex] 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.
 

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