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Sagot :
Certainly! Let's analyze the given function to identify its asymptotes:
[tex]\[ f(x) = \frac{5x^2 + 4}{x} \][/tex]
### Part 1: Vertical Asymptotes
Vertical asymptotes occur where the denominator of the function is equal to zero (and the numerator is not zero at those points because we'd then have a hole).
The denominator of [tex]\( f(x) \)[/tex] is [tex]\( x \)[/tex].
To find the vertical asymptote:
[tex]\[ x = 0 \][/tex]
So the equation of the vertical asymptote is:
[tex]\[ \boxed{x = 0} \][/tex]
### Part 2: Horizontal and Slant Asymptotes
#### Horizontal Asymptotes
Horizontal asymptotes describe the behavior of the function as [tex]\( x \)[/tex] approaches infinity ([tex]\(\infty\)[/tex]) or negative infinity ([tex]\(-\infty\)[/tex]).
To find the horizontal asymptote, we evaluate the limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches [tex]\(\infty\)[/tex]:
[tex]\[ \lim_{{x \to \infty}} \frac{5x^2 + 4}{x} \][/tex]
This essentially looks at the highest degree terms in the numerator and the denominator:
[tex]\[ \lim_{{x \to \infty}} \frac{5x^2}{x} = \lim_{{x \to \infty}} 5x = \infty \][/tex]
So, there is no horizontal asymptote since [tex]\( \lim_{{x \to \infty}} f(x) \)[/tex] does not result in a finite value.
Thus, the equation for the horizontal asymptote is:
[tex]\[ \boxed{\text{None}} \][/tex]
#### Slant (Oblique) Asymptotes
A slant asymptote occurs if the degree of the numerator is exactly one more than the degree of the denominator. In this case, the degree of the numerator ([tex]\(5x^2 + 4\)[/tex]) is 2, and the degree of the denominator ([tex]\(x\)[/tex]) is 1. Hence, we can find the slant asymptote.
To find the slant asymptote, we perform polynomial long division of [tex]\( \frac{5x^2 + 4}{x} \)[/tex]:
[tex]\[ \frac{5x^2 + 4}{x} = 5x + \frac{4}{x} \][/tex]
As [tex]\(x \to \infty\)[/tex], the term [tex]\(\frac{4}{x}\)[/tex] approaches 0. Therefore, the slant asymptote is:
[tex]\[ y = 5x \][/tex]
Hence, the equation of the slant asymptote is:
[tex]\[ \boxed{y = 5x} \][/tex]
In summary:
- The vertical asymptote is [tex]\( x = 0 \)[/tex].
- There is no horizontal asymptote.
- The slant asymptote is [tex]\( y = 5x \)[/tex].
[tex]\[ f(x) = \frac{5x^2 + 4}{x} \][/tex]
### Part 1: Vertical Asymptotes
Vertical asymptotes occur where the denominator of the function is equal to zero (and the numerator is not zero at those points because we'd then have a hole).
The denominator of [tex]\( f(x) \)[/tex] is [tex]\( x \)[/tex].
To find the vertical asymptote:
[tex]\[ x = 0 \][/tex]
So the equation of the vertical asymptote is:
[tex]\[ \boxed{x = 0} \][/tex]
### Part 2: Horizontal and Slant Asymptotes
#### Horizontal Asymptotes
Horizontal asymptotes describe the behavior of the function as [tex]\( x \)[/tex] approaches infinity ([tex]\(\infty\)[/tex]) or negative infinity ([tex]\(-\infty\)[/tex]).
To find the horizontal asymptote, we evaluate the limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches [tex]\(\infty\)[/tex]:
[tex]\[ \lim_{{x \to \infty}} \frac{5x^2 + 4}{x} \][/tex]
This essentially looks at the highest degree terms in the numerator and the denominator:
[tex]\[ \lim_{{x \to \infty}} \frac{5x^2}{x} = \lim_{{x \to \infty}} 5x = \infty \][/tex]
So, there is no horizontal asymptote since [tex]\( \lim_{{x \to \infty}} f(x) \)[/tex] does not result in a finite value.
Thus, the equation for the horizontal asymptote is:
[tex]\[ \boxed{\text{None}} \][/tex]
#### Slant (Oblique) Asymptotes
A slant asymptote occurs if the degree of the numerator is exactly one more than the degree of the denominator. In this case, the degree of the numerator ([tex]\(5x^2 + 4\)[/tex]) is 2, and the degree of the denominator ([tex]\(x\)[/tex]) is 1. Hence, we can find the slant asymptote.
To find the slant asymptote, we perform polynomial long division of [tex]\( \frac{5x^2 + 4}{x} \)[/tex]:
[tex]\[ \frac{5x^2 + 4}{x} = 5x + \frac{4}{x} \][/tex]
As [tex]\(x \to \infty\)[/tex], the term [tex]\(\frac{4}{x}\)[/tex] approaches 0. Therefore, the slant asymptote is:
[tex]\[ y = 5x \][/tex]
Hence, the equation of the slant asymptote is:
[tex]\[ \boxed{y = 5x} \][/tex]
In summary:
- The vertical asymptote is [tex]\( x = 0 \)[/tex].
- There is no horizontal asymptote.
- The slant asymptote is [tex]\( y = 5x \)[/tex].
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