Question

finding asymptotes in exercises 1–4, find all ve asymptotes of the graph of the function.

1. $f(x)=\frac{4}{x^2}$

answer

1. $f(x)=\frac{5 + x}{5 - x}$
2. $f(x)=\frac{x^3}{x^2 - 1}$

answer

1. $g(x)=\frac{x^2}{x^2 - 4}$

Explanation:

Let's solve each function for its vertical asymptotes (I'll assume "ve" in the problem is a typo for "vertical"). A vertical asymptote occurs where the function is undefined (denominator is zero) and the limit as \( x \) approaches that value is \( \pm\infty \).

1. \( f(x) = \frac{4}{x^2} \)

Step 1: Find where the denominator is zero.

The denominator is \( x^2 \). Set \( x^2 = 0 \):
\( x = 0 \).

Step 2: Check the limit as \( x \to 0 \).

As \( x \to 0^+ \) (approaches 0 from the right), \( x^2 \to 0^+ \), so \( \frac{4}{x^2} \to +\infty \).
As \( x \to 0^- \) (approaches 0 from the left), \( x^2 \to 0^+ \) (since squaring a negative number gives a positive), so \( \frac{4}{x^2} \to +\infty \).
Thus, \( x = 0 \) is a vertical asymptote.

2. \( f(x) = \frac{5 + x}{5 - x} \)

Step 1: Find where the denominator is zero.

The denominator is \( 5 - x \). Set \( 5 - x = 0 \):
\( x = 5 \).

Step 2: Check the limit as \( x \to 5 \).

As \( x \to 5^+ \) (approaches 5 from the right), \( 5 - x \to 0^- \) (negative), so \( \frac{5 + x}{5 - x} \to \frac{10}{0^-} = -\infty \).
As \( x \to 5^- \) (approaches 5 from the left), \( 5 - x \to 0^+ \) (positive), so \( \frac{5 + x}{5 - x} \to \frac{10}{0^+} = +\infty \).
Thus, \( x = 5 \) is a vertical asymptote.

3. \( f(x) = \frac{x^3}{x^2 - 1} \)

Step 1: Factor the denominator.

\( x^2 - 1 = (x - 1)(x + 1) \) (difference of squares).

Step 2: Find where the denominator is zero.

Set \( (x - 1)(x + 1) = 0 \):
\( x = 1 \) or \( x = -1 \).

Step 3: Check limits for \( x = 1 \) and \( x = -1 \).

• For \( x = 1 \):

As \( x \to 1^+ \), \( x^2 - 1 \to 0^+ \), \( x^3 \to 1 \), so \( \frac{x^3}{x^2 - 1} \to \frac{1}{0^+} = +\infty \).
As \( x \to 1^- \), \( x^2 - 1 \to 0^- \), \( x^3 \to 1 \), so \( \frac{x^3}{x^2 - 1} \to \frac{1}{0^-} = -\infty \).

• For \( x = -1 \):

As \( x \to -1^+ \), \( x^2 - 1 \to 0^- \), \( x^3 \to -1 \), so \( \frac{x^3}{x^2 - 1} \to \frac{-1}{0^-} = +\infty \).
As \( x \to -1^- \), \( x^2 - 1 \to 0^+ \), \( x^3 \to -1 \), so \( \frac{x^3}{x^2 - 1} \to \frac{-1}{0^+} = -\infty \).

Thus, vertical asymptotes at \( x = 1 \) and \( x = -1 \).

4. \( g(x) = \frac{x^2}{x^2 - 4} \)

Step 1: Factor the denominator.

\( x^2 - 4 = (x - 2)(x + 2) \) (difference of squares).

Step 2: Find where the denominator is zero.

Set \( (x - 2)(x + 2) = 0 \):
\( x = 2 \) or \( x = -2 \).

Step 3: Check limits for \( x = 2 \) and \( x = -2 \).

• For \( x = 2 \):

As \( x \to 2^+ \), \( x^2 - 4 \to 0^+ \), \( x^2 \to 4 \), so \( \frac{x^2}{x^2 - 4} \to \frac{4}{0^+} = +\infty \).
As \( x \to 2^- \), \( x^2 - 4 \to 0^- \), \( x^2 \to 4 \), so \( \frac{x^2}{x^2 - 4} \to \frac{4}{0^-} = -\infty \).

• For \( x = -2 \):

As \( x \to -2^+ \), \( x^2 - 4 \to 0^- \), \( x^2 \to 4 \), so \( \frac{x^2}{x^2 - 4} \to \frac{4}{0^-} = -\infty \).
As \( x \to -2^- \), \( x^2 - 4 \to 0^+ \), \( x^2 \to 4 \), so \( \frac{x^2}{x^2 - 4} \to \frac{4}{0^+} = +\infty \).

Thus, vertical asymptotes at \( x = 2 \) and \( x = -2 \).

Final Answers:

1. \( \boldsymbol{x = 0} \)
2. \( \boldsymbol{x = 5} \)
3. \( \boldsymbol{x = 1} \) and \( \boldsymbol{x = -1} \)
4. \( \boldsymbol{x = 2} \) and \( \boldsymbol{x = -2} \)

Answer:

Let's solve each function for its vertical asymptotes (I'll assume "ve" in the problem is a typo for "vertical"). A vertical asymptote occurs where the function is undefined (denominator is zero) and the limit as \( x \) approaches that value is \( \pm\infty \).

1. \( f(x) = \frac{4}{x^2} \)

Step 1: Find where the denominator is zero.

The denominator is \( x^2 \). Set \( x^2 = 0 \):
\( x = 0 \).

Step 2: Check the limit as \( x \to 0 \).

As \( x \to 0^+ \) (approaches 0 from the right), \( x^2 \to 0^+ \), so \( \frac{4}{x^2} \to +\infty \).
As \( x \to 0^- \) (approaches 0 from the left), \( x^2 \to 0^+ \) (since squaring a negative number gives a positive), so \( \frac{4}{x^2} \to +\infty \).
Thus, \( x = 0 \) is a vertical asymptote.

2. \( f(x) = \frac{5 + x}{5 - x} \)

Step 1: Find where the denominator is zero.

The denominator is \( 5 - x \). Set \( 5 - x = 0 \):
\( x = 5 \).

Step 2: Check the limit as \( x \to 5 \).

As \( x \to 5^+ \) (approaches 5 from the right), \( 5 - x \to 0^- \) (negative), so \( \frac{5 + x}{5 - x} \to \frac{10}{0^-} = -\infty \).
As \( x \to 5^- \) (approaches 5 from the left), \( 5 - x \to 0^+ \) (positive), so \( \frac{5 + x}{5 - x} \to \frac{10}{0^+} = +\infty \).
Thus, \( x = 5 \) is a vertical asymptote.

3. \( f(x) = \frac{x^3}{x^2 - 1} \)

Step 1: Factor the denominator.

\( x^2 - 1 = (x - 1)(x + 1) \) (difference of squares).

Step 2: Find where the denominator is zero.

Set \( (x - 1)(x + 1) = 0 \):
\( x = 1 \) or \( x = -1 \).

Step 3: Check limits for \( x = 1 \) and \( x = -1 \).

• For \( x = 1 \):

As \( x \to 1^+ \), \( x^2 - 1 \to 0^+ \), \( x^3 \to 1 \), so \( \frac{x^3}{x^2 - 1} \to \frac{1}{0^+} = +\infty \).
As \( x \to 1^- \), \( x^2 - 1 \to 0^- \), \( x^3 \to 1 \), so \( \frac{x^3}{x^2 - 1} \to \frac{1}{0^-} = -\infty \).

• For \( x = -1 \):

As \( x \to -1^+ \), \( x^2 - 1 \to 0^- \), \( x^3 \to -1 \), so \( \frac{x^3}{x^2 - 1} \to \frac{-1}{0^-} = +\infty \).
As \( x \to -1^- \), \( x^2 - 1 \to 0^+ \), \( x^3 \to -1 \), so \( \frac{x^3}{x^2 - 1} \to \frac{-1}{0^+} = -\infty \).

Thus, vertical asymptotes at \( x = 1 \) and \( x = -1 \).

4. \( g(x) = \frac{x^2}{x^2 - 4} \)

Step 1: Factor the denominator.

\( x^2 - 4 = (x - 2)(x + 2) \) (difference of squares).

Step 2: Find where the denominator is zero.

Set \( (x - 2)(x + 2) = 0 \):
\( x = 2 \) or \( x = -2 \).

Step 3: Check limits for \( x = 2 \) and \( x = -2 \).

• For \( x = 2 \):

As \( x \to 2^+ \), \( x^2 - 4 \to 0^+ \), \( x^2 \to 4 \), so \( \frac{x^2}{x^2 - 4} \to \frac{4}{0^+} = +\infty \).
As \( x \to 2^- \), \( x^2 - 4 \to 0^- \), \( x^2 \to 4 \), so \( \frac{x^2}{x^2 - 4} \to \frac{4}{0^-} = -\infty \).

• For \( x = -2 \):

As \( x \to -2^+ \), \( x^2 - 4 \to 0^- \), \( x^2 \to 4 \), so \( \frac{x^2}{x^2 - 4} \to \frac{4}{0^-} = -\infty \).
As \( x \to -2^- \), \( x^2 - 4 \to 0^+ \), \( x^2 \to 4 \), so \( \frac{x^2}{x^2 - 4} \to \frac{4}{0^+} = +\infty \).

Thus, vertical asymptotes at \( x = 2 \) and \( x = -2 \).

Final Answers:

1. \( \boldsymbol{x = 0} \)
2. \( \boldsymbol{x = 5} \)
3. \( \boldsymbol{x = 1} \) and \( \boldsymbol{x = -1} \)
4. \( \boldsymbol{x = 2} \) and \( \boldsymbol{x = -2} \)