Answer :
To determine which function [tex]\( f \)[/tex] has a vertical asymptote at [tex]\( x = 3 \)[/tex] and an oblique asymptote, let's examine the given options one by one:
1. [tex]\( f(m) = \frac{a - 3}{p_1 - 30 - 1} \)[/tex]
- The variables used here are [tex]\(m\)[/tex] and [tex]\(p_1\)[/tex], not [tex]\(x\)[/tex]. Since we need a function of [tex]\(x\)[/tex], this option is not valid.
2. [tex]\( f(x) = \frac{a + 3}{x - 2\pi - 7} \)[/tex]
- For a vertical asymptote at [tex]\( x = 3 \)[/tex], the denominator should be zero when [tex]\( x = 3 \)[/tex].
- Substituting [tex]\( x = 3 \)[/tex] in [tex]\( x - 2\pi - 7 \)[/tex] gives [tex]\( 3 - 2\pi - 7 = 3 - 2\pi - 7 \neq 0 \)[/tex].
- Therefore, this function does not have a vertical asymptote at [tex]\( x = 3 \)[/tex].
3. [tex]\( f(x) = \frac{x^3 - 3 - 1}{x + 3} \)[/tex]
- For a vertical asymptote at [tex]\( x = 3 \)[/tex], the denominator should be zero when [tex]\( x = 3 \)[/tex].
- Substituting [tex]\( x = 3 \)[/tex] in [tex]\( x + 3 \)[/tex] gives [tex]\( 3 + 3 = 6 \neq 0 \)[/tex].
- Therefore, this function does not have a vertical asymptote at [tex]\( x = 3 \)[/tex].
4. [tex]\( f(x) = \frac{x^3 - 3 - 1}{x - 3} \)[/tex]
- For a vertical asymptote at [tex]\( x = 3 \)[/tex], the denominator should be zero when [tex]\( x = 3 \)[/tex].
- Substituting [tex]\( x = 3 \)[/tex] in [tex]\( x - 3 \)[/tex] gives [tex]\( 3 - 3 = 0 \)[/tex], satisfying the condition for a vertical asymptote.
- Next, we need to check if this function has an oblique asymptote. Oblique asymptotes occur when the degree of the numerator is exactly one higher than the degree of the denominator.
- The numerator [tex]\( x^3 - 4 \)[/tex] is a polynomial of degree 3.
- The denominator [tex]\( x - 3 \)[/tex] is a polynomial of degree 1.
- The degree of the numerator (3) is one higher than the degree of the denominator (1), so this function has an oblique asymptote.
Given these observations, option 4 correctly represents a function [tex]\( f \)[/tex] with a vertical asymptote at [tex]\( x = 3 \)[/tex] and an oblique asymptote. Thus, the answer is:
[tex]\[ \boxed{4} \][/tex]
1. [tex]\( f(m) = \frac{a - 3}{p_1 - 30 - 1} \)[/tex]
- The variables used here are [tex]\(m\)[/tex] and [tex]\(p_1\)[/tex], not [tex]\(x\)[/tex]. Since we need a function of [tex]\(x\)[/tex], this option is not valid.
2. [tex]\( f(x) = \frac{a + 3}{x - 2\pi - 7} \)[/tex]
- For a vertical asymptote at [tex]\( x = 3 \)[/tex], the denominator should be zero when [tex]\( x = 3 \)[/tex].
- Substituting [tex]\( x = 3 \)[/tex] in [tex]\( x - 2\pi - 7 \)[/tex] gives [tex]\( 3 - 2\pi - 7 = 3 - 2\pi - 7 \neq 0 \)[/tex].
- Therefore, this function does not have a vertical asymptote at [tex]\( x = 3 \)[/tex].
3. [tex]\( f(x) = \frac{x^3 - 3 - 1}{x + 3} \)[/tex]
- For a vertical asymptote at [tex]\( x = 3 \)[/tex], the denominator should be zero when [tex]\( x = 3 \)[/tex].
- Substituting [tex]\( x = 3 \)[/tex] in [tex]\( x + 3 \)[/tex] gives [tex]\( 3 + 3 = 6 \neq 0 \)[/tex].
- Therefore, this function does not have a vertical asymptote at [tex]\( x = 3 \)[/tex].
4. [tex]\( f(x) = \frac{x^3 - 3 - 1}{x - 3} \)[/tex]
- For a vertical asymptote at [tex]\( x = 3 \)[/tex], the denominator should be zero when [tex]\( x = 3 \)[/tex].
- Substituting [tex]\( x = 3 \)[/tex] in [tex]\( x - 3 \)[/tex] gives [tex]\( 3 - 3 = 0 \)[/tex], satisfying the condition for a vertical asymptote.
- Next, we need to check if this function has an oblique asymptote. Oblique asymptotes occur when the degree of the numerator is exactly one higher than the degree of the denominator.
- The numerator [tex]\( x^3 - 4 \)[/tex] is a polynomial of degree 3.
- The denominator [tex]\( x - 3 \)[/tex] is a polynomial of degree 1.
- The degree of the numerator (3) is one higher than the degree of the denominator (1), so this function has an oblique asymptote.
Given these observations, option 4 correctly represents a function [tex]\( f \)[/tex] with a vertical asymptote at [tex]\( x = 3 \)[/tex] and an oblique asymptote. Thus, the answer is:
[tex]\[ \boxed{4} \][/tex]
