Find an even function that has vertical asymptotes x=1 and x=-2

Answer 1

If it has vertical asymptotes at the points x = 1 and x = -2, then we can write a minimialistic function as...

f(x) = A / [ (x-1)(x+2) ] where A is a constant ( A ≠ 0 )

If it were an even function then f(-x) = f(x).

Plugging in -x for there argument to f we get ...

f(-x) = A / [ (-x-1)(-x+2) ] = A / [ (-1)(x+1)(-1)(x-2) ] = A / [ (x+1)(x-2) ] ≠ f(x)

So we see that the asymptote at x = 1 is causing problems with evenness. We need to provide a mirror side which allows for symmetry then...

So we augment f(x) by --> x+1...

f(x) = A/ [ (x-1)(x+1)(x+2)(x-2) ]

Check again for evenness property...

f(-x) = A / [ (-x-1)(-x+1)(-x+2)(-x-2) ] = A / [ (-1)(x+1)(-1)(x-1)(-1)(x-2)(-1)(x+2) ] = A / [ (x+1)(x-1)(x-2)(x+2) ] = f(x)

So we have found a minimal function which satisfies the requirements for asymptotes and is an even function.

Answer 2

Michael K's answer is the same I'd have given.