Answer :
To analyze the function [tex]\( f(x) = \frac{2x + 24}{x^2 + 4x - 96} \)[/tex], let's go through each statement step-by-step based on the given characteristics of the function:
1. Vertical Asymptotes:
Vertical asymptotes occur where the denominator is zero, but the numerator is not zero at those points.
- Factorize the denominator [tex]\( x^2 + 4x - 96 = 0 \)[/tex]:
[tex]\[ x = -12 \quad \text{or} \quad x = 8 \][/tex]
Thus, the vertical asymptotes are at [tex]\( x = -12 \)[/tex] and [tex]\( x = 8 \)[/tex].
2. Hole in the Graph:
A hole occurs where both the numerator and denominator are zero for the same value of [tex]\( x \)[/tex].
- Solve [tex]\( 2x + 24 = 0 \)[/tex]:
[tex]\[ x = -12 \][/tex]
Therefore, there is a hole at [tex]\( x = -12 \)[/tex].
3. Horizontal Asymptote:
To determine if the function has a horizontal asymptote, we compare the degrees of the numerator and denominator.
- The degree of the numerator [tex]\( 2x + 24 \)[/tex] is 1.
- The degree of the denominator [tex]\( x^2 + 4x - 96 \)[/tex] is 2.
Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is:
[tex]\[ y = 0 \][/tex]
4. X-Intercept:
The x-intercept occurs where the numerator is zero (i.e., the function equals zero).
- Solve [tex]\( 2x + 24 = 0 \)[/tex]:
[tex]\[ x = -24 \][/tex]
Thus, the x-intercept is [tex]\((-12, 0)\)[/tex].
5. Increasing or Decreasing from [tex]\( (8, \infty) \)[/tex]:
To determine whether [tex]\( f(x) \)[/tex] is increasing or decreasing on the interval [tex]\( (8, \infty) \)[/tex], analyze the derivative [tex]\( f'(x) \)[/tex] over that interval.
- Evaluate the derivative at a point greater than 8, for example at [tex]\( x = 9 \)[/tex]:
[tex]\[ f'(9) < 0 \][/tex]
Since the derivative is negative, the function is decreasing on the interval [tex]\( (8, \infty) \)[/tex].
Summarizing the findings:
- [tex]\( f(x) \)[/tex] has a vertical asymptote at [tex]\( x = 8 \)[/tex]. ✓
- [tex]\( f(x) \)[/tex] has a vertical asymptote at [tex]\( x = -12 \)[/tex]. (Corrected answer)
- [tex]\( f(x) \)[/tex] has a hole at [tex]\( x = -12 \)[/tex]. ✓
- [tex]\( f(x) \)[/tex] has a horizontal asymptote at [tex]\( y = 0 \)[/tex]. ✓
- [tex]\( f(x) \)[/tex] has an [tex]\( x \)[/tex]-intercept at [tex]\( (-12, 0) \)[/tex]. ✓
- [tex]\( f(x) \)[/tex] is decreasing from [tex]\( (8, \infty) \)[/tex]. ✓
The statements that apply to the function are:
- [tex]\( f(x) \)[/tex] has a vertical asymptote at [tex]\( x = 8 \)[/tex].
- [tex]\( f(x) \)[/tex] has a hole at [tex]\( x = -12 \)[/tex].
- [tex]\( f(x) \)[/tex] has a horizontal asymptote at [tex]\( y = 0 \)[/tex].
- [tex]\( f(x) \)[/tex] is decreasing from [tex]\( (8, \infty) \)[/tex].
1. Vertical Asymptotes:
Vertical asymptotes occur where the denominator is zero, but the numerator is not zero at those points.
- Factorize the denominator [tex]\( x^2 + 4x - 96 = 0 \)[/tex]:
[tex]\[ x = -12 \quad \text{or} \quad x = 8 \][/tex]
Thus, the vertical asymptotes are at [tex]\( x = -12 \)[/tex] and [tex]\( x = 8 \)[/tex].
2. Hole in the Graph:
A hole occurs where both the numerator and denominator are zero for the same value of [tex]\( x \)[/tex].
- Solve [tex]\( 2x + 24 = 0 \)[/tex]:
[tex]\[ x = -12 \][/tex]
Therefore, there is a hole at [tex]\( x = -12 \)[/tex].
3. Horizontal Asymptote:
To determine if the function has a horizontal asymptote, we compare the degrees of the numerator and denominator.
- The degree of the numerator [tex]\( 2x + 24 \)[/tex] is 1.
- The degree of the denominator [tex]\( x^2 + 4x - 96 \)[/tex] is 2.
Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is:
[tex]\[ y = 0 \][/tex]
4. X-Intercept:
The x-intercept occurs where the numerator is zero (i.e., the function equals zero).
- Solve [tex]\( 2x + 24 = 0 \)[/tex]:
[tex]\[ x = -24 \][/tex]
Thus, the x-intercept is [tex]\((-12, 0)\)[/tex].
5. Increasing or Decreasing from [tex]\( (8, \infty) \)[/tex]:
To determine whether [tex]\( f(x) \)[/tex] is increasing or decreasing on the interval [tex]\( (8, \infty) \)[/tex], analyze the derivative [tex]\( f'(x) \)[/tex] over that interval.
- Evaluate the derivative at a point greater than 8, for example at [tex]\( x = 9 \)[/tex]:
[tex]\[ f'(9) < 0 \][/tex]
Since the derivative is negative, the function is decreasing on the interval [tex]\( (8, \infty) \)[/tex].
Summarizing the findings:
- [tex]\( f(x) \)[/tex] has a vertical asymptote at [tex]\( x = 8 \)[/tex]. ✓
- [tex]\( f(x) \)[/tex] has a vertical asymptote at [tex]\( x = -12 \)[/tex]. (Corrected answer)
- [tex]\( f(x) \)[/tex] has a hole at [tex]\( x = -12 \)[/tex]. ✓
- [tex]\( f(x) \)[/tex] has a horizontal asymptote at [tex]\( y = 0 \)[/tex]. ✓
- [tex]\( f(x) \)[/tex] has an [tex]\( x \)[/tex]-intercept at [tex]\( (-12, 0) \)[/tex]. ✓
- [tex]\( f(x) \)[/tex] is decreasing from [tex]\( (8, \infty) \)[/tex]. ✓
The statements that apply to the function are:
- [tex]\( f(x) \)[/tex] has a vertical asymptote at [tex]\( x = 8 \)[/tex].
- [tex]\( f(x) \)[/tex] has a hole at [tex]\( x = -12 \)[/tex].
- [tex]\( f(x) \)[/tex] has a horizontal asymptote at [tex]\( y = 0 \)[/tex].
- [tex]\( f(x) \)[/tex] is decreasing from [tex]\( (8, \infty) \)[/tex].
