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Sagot :
Let's analyze the rational function [tex]\( y = \frac{9x^2 + 81x}{x^3 + 8x^2 - 9x} \)[/tex].
### Step-by-Step Solution:
1. Finding the [tex]\(x\)[/tex]-intercepts:
The [tex]\(x\)[/tex]-intercepts can be found by setting the numerator equal to zero and solving for [tex]\(x\)[/tex].
Numerator:
[tex]\[ 9x^2 + 81x = 0 \][/tex]
Factor out the common factor:
[tex]\[ 9x(x + 9) = 0 \][/tex]
So, the [tex]\(x\)[/tex]-intercepts are:
[tex]\[ x = -9 \quad \text{and} \quad x = 0 \][/tex]
2. Finding the horizontal asymptote:
Compare the degrees of the numerator and the denominator.
- The degree of the numerator [tex]\(9x^2 + 81x\)[/tex] is 2.
- The degree of the denominator [tex]\(x^3 + 8x^2 - 9x\)[/tex] is 3.
Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is:
[tex]\[ y = 0 \][/tex]
3. Finding the vertical asymptotes:
The vertical asymptotes occur where the denominator is equal to zero and the numerator is not zero at those values. Set the denominator equal to zero and solve for [tex]\(x\)[/tex].
Denominator:
[tex]\[ x^3 + 8x^2 - 9x = 0 \][/tex]
Factor out the common factor:
[tex]\[ x(x^2 + 8x - 9) = 0 \][/tex]
Solve the factored equation:
[tex]\[ x = 0 \quad \text{or} \quad x^2 + 8x - 9 = 0 \][/tex]
Solve the quadratic equation:
[tex]\[ x = \frac{-8 \pm \sqrt{64 + 36}}{2} = \frac{-8 \pm \sqrt{100}}{2} = \frac{-8 \pm 10}{2} \][/tex]
Therefore:
[tex]\[ x = 1 \quad \text{and} \quad x = -9 \][/tex]
So the vertical asymptotes are:
[tex]\[ x = -9, \quad x = 0, \quad \text{and} \quad x = 1 \][/tex]
4. Finding holes in the graph:
Holes occur where there are common factors in both the numerator and the denominator. The common factor found is [tex]\(x(x + 9)\)[/tex].
Since the common factors are [tex]\(x = -9\)[/tex] and [tex]\(x = 0\)[/tex], there are holes at these points.
5. Counting the number of holes:
There are exactly:
[tex]\[ 2 \quad \text{holes (at} \quad x = -9 \quad \text{and} \quad x = 0) \][/tex]
### Summary:
- [tex]\(x\)[/tex]-intercepts: [tex]\( x = -9 \)[/tex] and [tex]\( x = 0 \)[/tex]
- Horizontal asymptote: [tex]\( y = 0 \)[/tex]
- Vertical asymptotes: [tex]\( x = -9 \)[/tex], [tex]\( x = 0 \)[/tex], and [tex]\( x = 1 \)[/tex]
- Holes: [tex]\( x = -9 \)[/tex] and [tex]\( x = 0 \)[/tex]
There are [tex]\( \boxed{2} \)[/tex] holes in the graph of the given rational function.
### Step-by-Step Solution:
1. Finding the [tex]\(x\)[/tex]-intercepts:
The [tex]\(x\)[/tex]-intercepts can be found by setting the numerator equal to zero and solving for [tex]\(x\)[/tex].
Numerator:
[tex]\[ 9x^2 + 81x = 0 \][/tex]
Factor out the common factor:
[tex]\[ 9x(x + 9) = 0 \][/tex]
So, the [tex]\(x\)[/tex]-intercepts are:
[tex]\[ x = -9 \quad \text{and} \quad x = 0 \][/tex]
2. Finding the horizontal asymptote:
Compare the degrees of the numerator and the denominator.
- The degree of the numerator [tex]\(9x^2 + 81x\)[/tex] is 2.
- The degree of the denominator [tex]\(x^3 + 8x^2 - 9x\)[/tex] is 3.
Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is:
[tex]\[ y = 0 \][/tex]
3. Finding the vertical asymptotes:
The vertical asymptotes occur where the denominator is equal to zero and the numerator is not zero at those values. Set the denominator equal to zero and solve for [tex]\(x\)[/tex].
Denominator:
[tex]\[ x^3 + 8x^2 - 9x = 0 \][/tex]
Factor out the common factor:
[tex]\[ x(x^2 + 8x - 9) = 0 \][/tex]
Solve the factored equation:
[tex]\[ x = 0 \quad \text{or} \quad x^2 + 8x - 9 = 0 \][/tex]
Solve the quadratic equation:
[tex]\[ x = \frac{-8 \pm \sqrt{64 + 36}}{2} = \frac{-8 \pm \sqrt{100}}{2} = \frac{-8 \pm 10}{2} \][/tex]
Therefore:
[tex]\[ x = 1 \quad \text{and} \quad x = -9 \][/tex]
So the vertical asymptotes are:
[tex]\[ x = -9, \quad x = 0, \quad \text{and} \quad x = 1 \][/tex]
4. Finding holes in the graph:
Holes occur where there are common factors in both the numerator and the denominator. The common factor found is [tex]\(x(x + 9)\)[/tex].
Since the common factors are [tex]\(x = -9\)[/tex] and [tex]\(x = 0\)[/tex], there are holes at these points.
5. Counting the number of holes:
There are exactly:
[tex]\[ 2 \quad \text{holes (at} \quad x = -9 \quad \text{and} \quad x = 0) \][/tex]
### Summary:
- [tex]\(x\)[/tex]-intercepts: [tex]\( x = -9 \)[/tex] and [tex]\( x = 0 \)[/tex]
- Horizontal asymptote: [tex]\( y = 0 \)[/tex]
- Vertical asymptotes: [tex]\( x = -9 \)[/tex], [tex]\( x = 0 \)[/tex], and [tex]\( x = 1 \)[/tex]
- Holes: [tex]\( x = -9 \)[/tex] and [tex]\( x = 0 \)[/tex]
There are [tex]\( \boxed{2} \)[/tex] holes in the graph of the given rational function.
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