Question 927655
Let {{{f(x) = (x^3 - 9x)/(4x^2 - 4x - 80)}}}



To prove an function is odd, we need to show that {{{f(-x) = -f(x)}}}. So let's first find {{{f(-x)}}}



{{{f(x) = (x^3 - 9x)/(4x^2 - 4x - 80)}}}



{{{f(-x) = ((-x)^3 - 9(-x))/(4(-x)^2 - 4(-x) - 80)}}}



{{{f(-x) = (-x^3 + 9)/(4x^2 + 4x - 80)}}}



Now let's find {{{-f(x)}}}



{{{f(x) = (x^3 - 9x)/(4x^2 - 4x - 80)}}}



{{{-f(x) = -(x^3 - 9x)/(4x^2 - 4x - 80)}}}



{{{-f(x) = (-x^3 + 9x)/(4x^2 - 4x - 80)}}}



So we can see that {{{f(-x)<>-f(x)}}} (notice how the +4x and -4x in the denominators don't match).



Therefore, {{{f(x)}}} is NOT an odd function.



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To show that an function is even, we need to show that {{{f(-x) = f(x)}}}



From the last part, we found that {{{f(-x) = (-x^3 + 9)/(4x^2 + 4x - 80)}}} but this is not equal to {{{f(x) = (x^3 - 9x)/(4x^2 - 4x - 80)}}}. 



The numerators don't match at all in terms of their signs and the denominators are off too a bit (again the +4x and -4x).



Therefore, {{{f(x)}}} is NOT an even function.



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Summary: Above we've shown that {{{f(x)}}} is neither even nor odd. 



Because of this, it does NOT have y axis symmetry (it's not even) and it does NOT have origin symmetry (it's not odd).



The graph visually sums it all up for us without having to say a single word. It looks like the graph has origin symmetry, but it's approximately symmetric for {{{-3<=x<=3}}}. Any x outside this interval has this symmetry breaking down.



{{{ graph( 500, 500, -10, 10, -10, 10, 0,(x^3 - 9x)/(4x^2 - 4x - 80)) }}}



From {{{x = -3}}} to {{{x = 3}}}, this graph looks pretty symmetric (with respect to the origin), but as x goes beyond +4, the curve does not go beyond -4 to reciprocate which is where the symmetry starts to break down.


A more basic and quick indication is to notice how the vertical asymptotes {{{x = -4}}} and {{{x = 5}}} are not equal in magnitude. Ie, {{{abs(-4) <> abs(5)}}}, so this also tells us f(x) does not have origin symmetry.


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