Question 1031765
(a)



{{{f(x) = (2x^2-1)/(x^2)}}}



{{{f(5) = (2(5)^2-1)/((5)^2)}}} Replace EVERY copy of x with 5



{{{f(5) = (2(25)-1)/(25)}}}



{{{f(5) = (50-1)/(25)}}}



{{{f(5) = 49/25}}}



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(b)



{{{f(x) = (2x^2-1)/(x^2)}}}



{{{f(-5) = (2(-5)^2-1)/((-5)^2)}}} Replace EVERY copy of x with -5



{{{f(-5) = (2(25)-1)/(25)}}} See note below



{{{f(-5) = (50-1)/(25)}}}



{{{f(-5) = 49/25}}}



Note: squaring something like -5 yields a positive result since (-5)^2 = (-5)*(-5) = +25



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(c)



{{{f(x) = (2x^2-1)/(x^2)}}}



{{{f(-x) = (2(-x)^2-1)/((-x)^2)}}} Replace EVERY copy of x with -x



Now because {{{(-x)^2 = x^2}}} (based on the note above) this means that we end up with 



{{{f(-x) = (2x^2-1)/(x^2)}}}



which is the same as the original function. Only the left side is different. Because {{{f(-x) = f(x)}}}, for all values of x in the domain of f(x), this means that f(x) is an even function. It is symmetric with respect to the y axis.