Question

Determine the equations of the vertical and horizontal asymptotes, if any, for h(x)= (x+1)^2/x^2-1

Answer 1

Answer:

ANSWER

Vertical asymptote:

x=1

Horizontal asymptote:

y=1

EXPLANATION

The given rational function is

h(x) = \frac{ {(x + 1)}^{2} }{ {x}^{2} - 1 }h(x)=

x

2

−1

(x+1)

2

h(x) = \frac{ {(x + 1)}^{2} }{ ({x} - 1)(x + 1)}h(x)=

(x−1)(x+1)

(x+1)

2

h(x) = \frac{ (x + 1)(x + 1) }{ ({x} - 1)(x + 1)}h(x)=

(x−1)(x+1)

(x+1)(x+1)

h(x) = \frac{ x + 1}{ {x} - 1}h(x)=

x−1

x+1

The vertical asymptote occurs at

{x} - 1 = 0x−1=0

x = 1x=1

The vertical asymptotes is x=1

The degree of the numerator is the same as the degree of the denominator.

The horizontal asymptote of such rational function is found by expressing the coefficient of the leading term in the numerator over that of the denominator.

y = \frac{1}{1}y=

1

1

y=1

Answer 2

Let

$f(x) = \frac{n(x)}{d(x)}$

where n(x) and d(x) are polynomials.

For vertical asymptote of f(x) :-

Set d(x) = 0 and find 'x'. If the numerator n(x) is not zero for this 'x', then it's a vertical asymptote, otherwise it is a hole in the equation.

For horizontal asymptote of f(x) :-

1) If degree of n(x) > degree of d(x), there is no horizontal asymptote (there is an oblique asymptote).

2) If degree of n(x) < degree of d(x), then x-axis (y = 0) is the horizontal asymptote.

3) If degree of n(x) = degree of d(x), then the horizontal asymptote is

$y = \frac{coefficient \: of \: highest \: power \: of \: 'x' \: in \: n(x)}{coefficient \: of \: highest \: power \: of \: 'x' \: in \: d(x)}$

Hope this helps...