Answer :
Certainly! Let's work through each part of the given function [tex]\( y = \frac{x^2 - 2x - 8}{x^2 + x - 20} \)[/tex].
### a. Determine the vertical asymptotes or "holes" if any.
To find vertical asymptotes or "holes," we must identify the values of [tex]\( x \)[/tex] that make the denominator zero. These are potential locations for vertical asymptotes unless they also make the numerator zero (in which case they are "holes").
1. Find the denominator zeros:
[tex]\[ x^2 + x - 20 = 0 \][/tex]
Factor the quadratic equation:
[tex]\[ (x + 5)(x - 4) = 0 \][/tex]
So, the solutions are:
[tex]\[ x = -5 \quad \text{and} \quad x = 4 \][/tex]
2. Determine if these values also make the numerator zero:
- Check [tex]\( x = -5 \)[/tex]:
[tex]\[ x^2 - 2x - 8 = (-5)^2 - 2(-5) - 8 = 25 + 10 - 8 = 27 \neq 0 \][/tex]
Thus, [tex]\( x = -5 \)[/tex] is a vertical asymptote.
- Check [tex]\( x = 4 \)[/tex]:
[tex]\[ x^2 - 2x - 8 = 4^2 - 2(4) - 8 = 16 - 8 - 8 = 0 \][/tex]
Thus, [tex]\( x = 4 \)[/tex] makes both the numerator and denominator zero and is therefore a "hole."
Result:
- Vertical asymptote: [tex]\( x = -5 \)[/tex]
- Hole: [tex]\( x = 4 \)[/tex]
### b. Determine the horizontal asymptotes if any.
Horizontal asymptotes are determined by comparing the degrees (highest powers) of the numerator and the denominator.
Both the numerator and the denominator are quadratic polynomials ([tex]\( x^2 \)[/tex]). Thus, we compare the leading coefficients (coefficients of [tex]\( x^2 \)[/tex]):
- Leading coefficient of numerator: 1
- Leading coefficient of denominator: 1
Since the degrees are the same, the horizontal asymptote is the ratio of the leading coefficients:
[tex]\[ y = \frac{1}{1} = 1 \][/tex]
Result:
- Horizontal asymptote: [tex]\( y = 1 \)[/tex]
### c. Determine the [tex]\( y \)[/tex]-intercept if one exists.
The [tex]\( y \)[/tex]-intercept occurs when [tex]\( x = 0 \)[/tex]. Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ y = \frac{0^2 - 2(0) - 8}{0^2 + 0 - 20} = \frac{-8}{-20} = \frac{2}{5} \][/tex]
Result:
- [tex]\( y \)[/tex]-intercept: [tex]\( y = \frac{2}{5} \)[/tex]
### d. Determine the [tex]\( x \)[/tex]-intercept(s) if any exist.
The [tex]\( x \)[/tex]-intercepts occur where the numerator is zero. Set the numerator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 - 2x - 8 = 0 \][/tex]
Factor the quadratic equation:
[tex]\[ (x - 4)(x + 2) = 0 \][/tex]
So, the solutions are:
[tex]\[ x = 4 \quad \text{and} \quad x = -2 \][/tex]
Result:
- [tex]\( x \)[/tex]-intercepts: [tex]\( x = -2 \)[/tex] and [tex]\( x = 4 \)[/tex]
### Summary:
- Vertical asymptote: [tex]\( x = -5 \)[/tex]
- Hole: [tex]\( x = 4 \)[/tex]
- Horizontal asymptote: [tex]\( y = 1 \)[/tex]
- [tex]\( y \)[/tex]-intercept: [tex]\( y = \frac{2}{5} \)[/tex]
- [tex]\( x \)[/tex]-intercepts: [tex]\( x = -2 \)[/tex] and [tex]\( x = 4 \)[/tex]
### a. Determine the vertical asymptotes or "holes" if any.
To find vertical asymptotes or "holes," we must identify the values of [tex]\( x \)[/tex] that make the denominator zero. These are potential locations for vertical asymptotes unless they also make the numerator zero (in which case they are "holes").
1. Find the denominator zeros:
[tex]\[ x^2 + x - 20 = 0 \][/tex]
Factor the quadratic equation:
[tex]\[ (x + 5)(x - 4) = 0 \][/tex]
So, the solutions are:
[tex]\[ x = -5 \quad \text{and} \quad x = 4 \][/tex]
2. Determine if these values also make the numerator zero:
- Check [tex]\( x = -5 \)[/tex]:
[tex]\[ x^2 - 2x - 8 = (-5)^2 - 2(-5) - 8 = 25 + 10 - 8 = 27 \neq 0 \][/tex]
Thus, [tex]\( x = -5 \)[/tex] is a vertical asymptote.
- Check [tex]\( x = 4 \)[/tex]:
[tex]\[ x^2 - 2x - 8 = 4^2 - 2(4) - 8 = 16 - 8 - 8 = 0 \][/tex]
Thus, [tex]\( x = 4 \)[/tex] makes both the numerator and denominator zero and is therefore a "hole."
Result:
- Vertical asymptote: [tex]\( x = -5 \)[/tex]
- Hole: [tex]\( x = 4 \)[/tex]
### b. Determine the horizontal asymptotes if any.
Horizontal asymptotes are determined by comparing the degrees (highest powers) of the numerator and the denominator.
Both the numerator and the denominator are quadratic polynomials ([tex]\( x^2 \)[/tex]). Thus, we compare the leading coefficients (coefficients of [tex]\( x^2 \)[/tex]):
- Leading coefficient of numerator: 1
- Leading coefficient of denominator: 1
Since the degrees are the same, the horizontal asymptote is the ratio of the leading coefficients:
[tex]\[ y = \frac{1}{1} = 1 \][/tex]
Result:
- Horizontal asymptote: [tex]\( y = 1 \)[/tex]
### c. Determine the [tex]\( y \)[/tex]-intercept if one exists.
The [tex]\( y \)[/tex]-intercept occurs when [tex]\( x = 0 \)[/tex]. Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ y = \frac{0^2 - 2(0) - 8}{0^2 + 0 - 20} = \frac{-8}{-20} = \frac{2}{5} \][/tex]
Result:
- [tex]\( y \)[/tex]-intercept: [tex]\( y = \frac{2}{5} \)[/tex]
### d. Determine the [tex]\( x \)[/tex]-intercept(s) if any exist.
The [tex]\( x \)[/tex]-intercepts occur where the numerator is zero. Set the numerator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 - 2x - 8 = 0 \][/tex]
Factor the quadratic equation:
[tex]\[ (x - 4)(x + 2) = 0 \][/tex]
So, the solutions are:
[tex]\[ x = 4 \quad \text{and} \quad x = -2 \][/tex]
Result:
- [tex]\( x \)[/tex]-intercepts: [tex]\( x = -2 \)[/tex] and [tex]\( x = 4 \)[/tex]
### Summary:
- Vertical asymptote: [tex]\( x = -5 \)[/tex]
- Hole: [tex]\( x = 4 \)[/tex]
- Horizontal asymptote: [tex]\( y = 1 \)[/tex]
- [tex]\( y \)[/tex]-intercept: [tex]\( y = \frac{2}{5} \)[/tex]
- [tex]\( x \)[/tex]-intercepts: [tex]\( x = -2 \)[/tex] and [tex]\( x = 4 \)[/tex]
