How many vertical asymptotes does the graph of f(x) have?

if f(x) is

____2x + 1______

x^3 + x^2 + 5x + 5

0, 1, 2, 3 or none of these

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  • 1 decade ago
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    ti has one, x = -1. because x^3+x^2+5x+5 = x^2(x+1)+5(x+1)= (x+1)(x^2+5)

    x+1 = 0, for x=-1

    x^2+5 >0 for any x. therefore, only 1

  • 1 decade ago

    You have to check when denominator = 0.

    x^3 + x^2 + 5x + 5 = 0

    x^2(x+1) + 5 (x+1) = 0

    (x^2 + 5) (x+1) = 0

    Denominator = 0 when x=-1.

    lim _ ____2x + 1______ = + ∞

    x>-1 x^3 + x^2 + 5x + 5

    lim ____2x + 1______ = - ∞

    x>-1+ x^3 + x^2 + 5x + 5

    when y=f(x)--->∞ you have a vertical asymptote.

    There is only one vertical asymptote: f(x) = -1.

  • 1 decade ago

    In this case, f(x) has a vertical asymptote when (x^3+x^2+5x+5)=0.

    Or, if you simplify, (x+1)(x^2+5)=0.

    This means that vertical asymptotes are at x+1=0 and x^2+5=0.

    x+1=0 => x=-1

    x^2+5=0 => x^2=-5 => (assuming you do not use complex numbers) => no solution from this.

    Or, in other words: there is 1 vertical asymptote, at x=-1.

    Good luck.

    Source(s): High-school Math
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