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Square root graph

  1. Jul 6, 2007 #1
    I'm confused on the graph: [tex] y=\sqrt{x}-4[/tex]

    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.
     
  2. jcsd
  3. Jul 6, 2007 #2
    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.
     
  4. Jul 6, 2007 #3

    cristo

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    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.
     
  5. Jul 6, 2007 #4
    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)?
     
  6. Jul 6, 2007 #5

    It would go back to the positive-x because it is a side-ways parabola, [tex]x=(y+4)^2[/tex].
     
    Last edited: Jul 6, 2007
  7. Jul 6, 2007 #6

    Kurdt

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    I think you're confused between the choice of the range for the function and simply having a graph of [itex]f(x)=-\sqrt{x}[/itex]. 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.
     
  8. Jul 6, 2007 #7
    Ah, I see. I remember doing side ways parabolas. [tex] (y-k)^{2}=4p(x+h) [/tex] right?
     
  9. Jul 6, 2007 #8

    cristo

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    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.
     
  10. Jul 6, 2007 #9
    So, how is a sideways parabola plotted? Or a circle? Is x^0.5 used? Or some other mathematical trickery?
     
  11. Jul 6, 2007 #10

    cristo

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    Neither a circle nor a sideways parabola are functions.
     
  12. Jul 6, 2007 #11
    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!
     
  13. Jul 6, 2007 #12

    cristo

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    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 [itex]\sqrt{x}[/itex], 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 [itex]f(x)=\sqrt{x}[/itex] it really means [itex]f(x)=+\sqrt{x}[/itex]-- i.e. the range is chosen to be the positive real numbers. This is a convention which makes the graph you draw a function.
     
  14. Jul 6, 2007 #13
    Ah, I understand now. Thanks :smile:

    And sorry to OP for threadjack.
     
    Last edited: Jul 6, 2007
  15. Jul 6, 2007 #14

    Kurdt

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    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.
     
  16. Jul 6, 2007 #15
    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?
     
  17. Jul 6, 2007 #16
    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.
     
  18. Jul 6, 2007 #17

    Kurdt

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    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.
     
  19. Jul 6, 2007 #18
    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?
     
  20. Jul 6, 2007 #19

    cristo

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    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.

    You can draw a graph of anything, but it is not necessarily a function. The graph of a circle is not a function.
     
  21. Jul 6, 2007 #20
    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??
     
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