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Finding Intercepts of an Equation

  1. Sep 2, 2014 #1

    Drakkith

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    1. The problem statement, all variables and given/known data
    Find the intercepts of the following equation: [itex]y=\frac{-x^3}{x^2-9}[/itex]


    2. Relevant equations

    To find the X-Intercept, set Y equal to zero and solve.
    To find the Y-Intercept, set X equal to zero and solve.

    3. The attempt at a solution

    Setting Y equal to zero gives me: [itex] 0=\frac{-x^3}{x^2-9}[/itex]
    This simplifies to: [itex] 0=\frac{-x^3}{(x+3)(x-3)}[/itex]

    And now I'm stuck. :uhh:
     
  2. jcsd
  3. Sep 2, 2014 #2

    Mentallic

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    The first thing you should remind yourself of is that we can't divide by 0. Clearly since the denominator is

    [tex]x^2-9=(x-3)(x+3)[/tex]

    and this quadratic is equal to 0 when [itex]x=\pm 3[/itex] hence our original equation does not exist at those x values.

    So now, we consider when x is any other value.

    If we have some fraction a/b for any numbers a and b where b is anything except 0, then when is this fraction equal to 0?
     
  4. Sep 2, 2014 #3

    Drakkith

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    Never?
     
  5. Sep 2, 2014 #4

    Mentallic

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    What does 0/5 equal to? 0/100? 0/k? for any k that isn't 0?
     
  6. Sep 2, 2014 #5

    Drakkith

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    Either zero or undefined. I'm guessing zero?
     
  7. Sep 3, 2014 #6

    Drakkith

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    Okay, I looked it up and it's definitely zero.
     
  8. Sep 3, 2014 #7

    Mark44

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    Yes, definitely. Zero divided by any nonzero number is zero.
     
  9. Sep 3, 2014 #8

    Drakkith

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    Yeah, I feel bad that I had to go look that up. I guess it's been a while since I did math. :frown:
     
  10. Sep 3, 2014 #9

    Mentallic

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    It's ok, that's what we're here for! :smile:

    So you can look at this problem in two ways, either notice that a fraction of the form a/b is equal to zero when the numerator is equal to 0, so in your equation

    [tex]y=\frac{-x^3}{x^2-9}[/tex]

    To find when y=0, i.e. to solve

    [tex]0=\frac{-x^3}{x^2-9}[/tex]

    we just need to find when the numerator -x3=0.

    Or the other way, simply multiply both sides of the equation by the denominator of the fraction.

    [tex]0\times (x^2-9) = \frac{-x^3}{x^2-9}\times (x^2-9)[/tex]

    Now on the RHS we can cancel the factor of x2-9 in both the numerator and denominator to end up with just -x^3, and on the LHS, 0 times anything is still 0, so of course again we end up solving 0=-x3.
     
  11. Sep 3, 2014 #10

    Drakkith

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    Hmmm. The only thing I can think of that makes -x3 zero is x = 0.

    So that would make the X intercept zero then, correct?
     
  12. Sep 3, 2014 #11

    Mentallic

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    Yes and yes.
     
  13. Sep 3, 2014 #12

    Drakkith

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    Okay, then if we set X = 0 to find the Y intercept: Y=0/-9, which would make Y=0.

    So the intercepts are (0,0)?
     
  14. Sep 3, 2014 #13

    Mentallic

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    Yes, but I wouldn't call it two intercepts :wink:

    Just a little extra in case you're curious:
    We didn't even need to check x=0 (the intercept of the y-axis) because we already have from the x-axis intercept that (0,0) is satisfied.

    The alternative doesn't always suffice though. If we first checked x=0 (the y-axis intercept) and found y=0 satisfied this, we would still need to solve for y=0 because we could possibly have more solutions for x such as in the case of quadratics.

    Now, the reason why we wouldn't need to check x=0 (as opposed needing to check y=0) is because y is a function of x. Functions have a very important property that for every x-value, there is only ever one y-value, so when we found that for x=0, y=0, we couldn't possibly have another value for y given x=0, so we didn't need to test x=0.
     
  15. Sep 3, 2014 #14

    Drakkith

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    Interesting. I didn't even realize that having the X-intercept as 0 meant that we also had the Y intercept. Makes sense. Thanks, Mentallic!
     
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