Question

1. vertical and horizontal asymptotes

find the vertical asymptotes for the following functions and also determine what value they approach as (x
ightarrowpminfty).

1. (f(x)=\frac{1}{x^{2}+4x - 21})
2. (f(x)=\frac{1}{x^{2}-1})
3. (f(x)=\frac{x}{3x^{2}+13x - 10})
4. (f(x)=\frac{x^{2}}{12x^{2}+5x - 2})

also determine if the function approaches (pminfty) as the vertical asymptote is approached from the right and from the left.

Explanation:

Step1: Recall vertical - asymptote condition

Vertical asymptotes occur where the denominator of a rational function is zero. For a rational function $y = \frac{f(x)}{g(x)}$, we set $g(x)=0$.

Step2: Recall horizontal - asymptote rules

1. If the degree of the numerator $n$ and the degree of the denominator $m$ of a rational function $y=\frac{a_nx^n+\cdots}{b_mx^m+\cdots}$:

• If $n < m$, the horizontal asymptote is $y = 0$.
• If $n=m$, the horizontal asymptote is $y=\frac{a_n}{b_m}$.
• If $n>m$, there is no horizontal asymptote.

Step3: Solve for vertical asymptotes of $f(x)=\frac{1}{x^{2}+4x - 21}$

Set the denominator equal to zero: $x^{2}+4x - 21=(x + 7)(x - 3)=0$. So $x=-7$ and $x = 3$ are the vertical asymptotes.
Since the degree of the numerator is $0$ and the degree of the denominator is $2$ ($n = 0,m = 2$), the horizontal asymptote is $y = 0$.

Step4: Solve for vertical asymptotes of $f(x)=\frac{1}{x^{2}-1}$

Set the denominator equal to zero: $x^{2}-1=(x + 1)(x - 1)=0$. So $x=-1$ and $x = 1$ are the vertical asymptotes.
Since the degree of the numerator is $0$ and the degree of the denominator is $2$ ($n = 0,m = 2$), the horizontal asymptote is $y = 0$.

Step5: Solve for vertical asymptotes of $f(x)=\frac{x}{3x^{2}+13x - 10}$

Set the denominator equal to zero: $3x^{2}+13x - 10=(3x - 2)(x+5)=0$. So $x=\frac{2}{3}$ and $x=-5$ are the vertical asymptotes.
Since the degree of the numerator is $1$ and the degree of the denominator is $2$ ($n = 1,m = 2$), the horizontal asymptote is $y = 0$.

Step6: Solve for vertical asymptotes of $f(x)=\frac{x^{2}}{12x^{2}+5x - 2}$

Set the denominator equal to zero: $12x^{2}+5x - 2=(4x - 1)(3x+2)=0$. So $x=\frac{1}{4}$ and $x=-\frac{2}{3}$ are the vertical asymptotes.
Since the degree of the numerator is $2$ and the degree of the denominator is $2$ ($n = 2,m = 2$), and $a_n = 1,b_m=12$, the horizontal asymptote is $y=\frac{1}{12}$.

1. For $f(x)=\frac{1}{x^{2}+4x - 21}$:

• Vertical asymptotes: $x=-7,x = 3$
• Horizontal asymptote: $y = 0$
• As $x

ightarrow - 7^{-},f(x)
ightarrow-\infty$; as $x
ightarrow - 7^{+},f(x)
ightarrow+\infty$; as $x
ightarrow3^{-},f(x)
ightarrow-\infty$; as $x
ightarrow3^{+},f(x)
ightarrow+\infty$

1. For $f(x)=\frac{1}{x^{2}-1}$:

• Vertical asymptotes: $x=-1,x = 1$
• Horizontal asymptote: $y = 0$
• As $x

ightarrow - 1^{-},f(x)
ightarrow-\infty$; as $x
ightarrow - 1^{+},f(x)
ightarrow+\infty$; as $x
ightarrow1^{-},f(x)
ightarrow-\infty$; as $x
ightarrow1^{+},f(x)
ightarrow+\infty$

1. For $f(x)=\frac{x}{3x^{2}+13x - 10}$:

• Vertical asymptotes: $x=\frac{2}{3},x=-5$
• Horizontal asymptote: $y = 0$
• As $x

ightarrow\frac{2}{3}^{-},f(x)
ightarrow-\infty$; as $x
ightarrow\frac{2}{3}^{+},f(x)
ightarrow+\infty$; as $x
ightarrow - 5^{-},f(x)
ightarrow+\infty$; as $x
ightarrow - 5^{+},f(x)
ightarrow-\infty$

1. For $f(x)=\frac{x^{2}}{12x^{2}+5x - 2}$:

• Vertical asymptotes: $x=\frac{1}{4},x=-\frac{2}{3}$
• Horizontal asymptote: $y=\frac{1}{12}$
• As $x

ightarrow\frac{1}{4}^{-},f(x)
ightarrow-\infty$; as $x
ightarrow\frac{1}{4}^{+},f(x)
ightarrow+\infty$; as $x
ightarrow-\frac{2}{3}^{-},f(x)
ightarrow+\infty$; as $x
ightarrow-\frac{2}{3}^{+},f(x)
ightarrow-\infty$

Answer:

1. $f(x)=\frac{1}{x^{2}+4x - 21}$: Vertical asymptotes $x=-7,x = 3$, Horizontal asymptote $y = 0$
2. $f(x)=\frac{1}{x^{2}-1}$: Vertical asymptotes $x=-1,x = 1$, Horizontal asymptote $y = 0$
3. $f(x)=\frac{x}{3x^{2}+13x - 10}$: Vertical asymptotes $x=\frac{2}{3},x=-5$, Horizontal asymptote $y = 0$
4. $f(x)=\frac{x^{2}}{12x^{2}+5x - 2}$: Vertical asymptotes $x=\frac{1}{4},x=-\frac{2}{3}$, Horizontal asymptote $y=\frac{1}{12}$