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Question 167400: Will you please help me.
Determine the equation of any vertical asymptotes and the value of x for any holes in the graph: f(x)=x^2-11x+18/x-2
I must show my work and I am lost
Thanks
Answer by gonzo(654)   (Show Source):
You can put this solution on YOUR website! NOTE:
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since i posted this result, i did some research and have determined that there is definitely an anomaly at x = 2 but am still hesitant to say that an asymptote exists at that point.
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a manual calculation of the equation without factoring out shows the following when x = 2.
(x^2 - 11x + 18) / (x-2) becomes the following after simplification:
2^2 - 11*2 + 18 / 0
4 - 22 + 18 / 0
-18 + 18 / 0
0/0
the solution of 0/0 is indeterminate.
if i factor and cancel out x-2 before solving for x = 2, i get a real answer (-7).
if i don't factor and cancel out x-2 before solving for x = 2, or even if i do factor but don't cancel out x-2, i get 0/0.
anything divided by 0 is undefined. the result is usually represented by infinity.
0 divided by anything is 0
0/0 becomes a special case.
some say 0/0 should equal 1 but this is proven to be in error, because the answer can be any value, rather than just a specific value.
case in point:
0/0 = 1 because 1 * 0 = 0, but anything times 0 = 0 so this is totally different from, say:
10/5 = 2 because 2 * 5 = 10. in this case the answer has to be 5 or the reverse operation doesn't work.
bottom line:
answer of division by 0 is indeterminate in this case.
does it create an asymptote?
based on this definition of asymptote:
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A line is asymptotic to a curve if the distance between the line and the curve tends to zero as the distance along the curve tends to infinity.
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i would have to say no, because the distance along the curve does not appear to tend towards infinity at that point.
it appears to tend towards 0, which makes sense, since the value at x=2 is 0/0.
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end of NOTE:
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a vertical asymptote would occur when your divisor is 0.
at that point the value of your equation goes to infinity which is undefined.
since your divisor is x-2, it would be 0 when x = 2 (2-2=0).
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your full equation is:
(x^2 - 11x + 18) / (x-2)
if you factor the top equation, you get:
(x-9)*(x-2) / (x-2)
the (x-2) would cancel out and your equation would become (x-9).
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here's a graph of your original equation:
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i would have expected the graph to go to infinity at x = 2 only it didn't.
i suspect that's because when i factored the top equation, the factor of (x-2) on top cancels out the factor of (x-2) on the bottom so that the resultant equation doesn't have any divisor of zero complications.
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to show you what i mean, i changed the top equation so that it doesn't neatly factor out the (x-2).
i changed it to x^2 - 11x + 27.
that factors out to (x-9)*(x-3) which would NOT cancel out the (x-2) of the denominator.
that graph looks like this:
look below the graph for final comments.
i'm inclined to say that your original equation does NOT have any vertical asymptotes and the reason for that is that the divisor cancels out and so is not a factor in the final equation.
this is certainly supported by the first graph which is the original equation, and by the second graph which is a modified equation where the divisor doesn't cancel out.
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i tried another online graphing calculator with the same result.
the original equation which was your problem has no holes in it with no vertical asymptotes.
the modified equation does.
since your problem is the original equation you were presented with, the answer is no vertical asymptotes.
the reason appears to be what i strongly suspected. that is that the denominator cancels out so it has no impact on the final output of the equation.
it's a surprise to me, but seems logical.
if you get a chance, send me the result of what your instructor says about this.
email: gonzo@gmx.us.
thanks.
hope this helped.
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