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function y=x^2-5x+4/x^2+a^2x+2a has 1 vertical asymptotesshow how calculate  a=?

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lightjazz | Student, Undergraduate | eNoter

Posted February 27, 2012 at 2:08 PM via web

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function y=x^2-5x+4/x^2+a^2x+2a has 1 vertical asymptotes

show how calculate  a=?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted February 27, 2012 at 2:24 PM (Answer #1)

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You need to remember that vertical asymptote comes from restrictions on domain of definition of function.

Hence, the zeroes of denominator of fraction become restrictions on domain.

The probelm provides the information that the function has one vertical asymptote, hence there must be a single restriction on domain, hence the denominator needs to have two equal roots.

The denominator is a quadratic and you should remember that a quadratic has two equal roots if `Delta = 0` .

You need to remember what `Delta`  is such that:

`Delta = n^2 - 4mq`  (m,n,q express the coefficients of quadratic `mx^2 + nx + q` )

Hence `m=1,n = a^2 , q = 2a`  such that:

`Delta = a^4 - 8a`

You need to solve `Delta = 0` , hence you need to substitute `a^4 - 8a`  for `Delta ` such that:

`a^4 - 8a = 0`

Factoring out a yields: `a(a^3 - 8) = 0`

You need to expand the difference of cubes `a^3 - 8`  such that:

a(a-2)(a^2 + 2a + 4) = 0

`a = 0; a - 2 = 0 =gt a = 2`

Notice that `a^2 + 2a + 4`  never cancels for real values of a, hence, evaluating the values of a for the function has one vertical asymptote yields `a=0; a=2` .

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