Respuesta :
Answer:
- Domain: All the real values except x = 2 and x = 4: R - {2, 4}
- Holes: x = 2
- VA, vertical asymptores: x = 4
- HA: horizontal asymptotes: there are not horizontal asymptotes
- OA: oblique asymptotes: x + 6 [note that OH does not stand for any known feature, and so it is understood that it was intended to write OA]
- Roots: x = 2
- Y-intercept: -1
Step-by-step explanation:
1. Given:
[tex]f(x)=\frac{x^3-8}{x^2-6x+8}[/tex]
- Note that the number 8 in the numerator is not part of the power.
- Type of function: rational function
2. Domain: is the set of x-values for which the function is defined.
The given function is defined for all x except those for which the denominator equals 0.
- Denominator: x² -6x + 8 = 0
- Solve for x:
Factor. (x - 4 )(x - 2) = 0
Zero product property: (x - 4) = 0 or (x - 2) = 0
x - 4 = 0 ⇒ x = 4
x - 2 = 0 ⇒ x = 2
- Domain:
All the real values except x = 2 and x = 4: x ∈ R / x ≠ 2 and x ≠ 4.
3. Holes:
The holes on the graph of a rational function are at those x-values for which both the numerator and denominator are zero.
- Find the values for which the numerator is zero:
Numerator: x³ - 8 = 0
Factor using difference of cubes property:
a³ - b³ = (a - b)(a² + ab + b²)
x³ - 8 = (x - 2)(x² + 2x + 4) = 0
Zero product property: (x - 2)(x² + 2x + 4) = 0
x - 2 = 0 ⇒ x = 2
x² + 2x + 4 = 0 (this has not real solution)
- The values for which the denominator is zero were determined above: x = 2 and x = 4.
- Conclusion: for x = 2 both numerator and denominator equal 0, so this is a hole.
4. VA: Vertical asymptotes.
The vertical asymptotes on the graph of a rational function are the vertical lines for which only the denominator (and not the numerator) equals zero.
- In the previous part it was determined that happens when x = 4.
5. HA: Horizontal asymptotes.
In rational functions, if the numerator is a higher degree polynomial than the denominator, there is no horizontal asymptote.
6. OA: oblique asymptotes
- Find the quotient and the remainder.
x + 6
_______________
x² - 6x + 8 ) x³ + 0x² + 0x - 8
- x³ + 6x² - 8x
___________
6 x² - 8x - 8
- 6x² + 36x - 48
_____________
28x - 56
Result: (x + 6) + (28x - 56) / (x² - 6x + 8)
- Find limit x → ∞
[tex]\lim_{x \to \infty}(x + 6) + \frac{28x-56}{x^2-6x+8}=x+6[/tex]
7. Roots:
Roots are the values for which f(x) = 0.
That happens when the numerator equals 0, and the denominator is not 0.
As determined earlier: x³ - 8 = 0 ⇒ x = 2.
8. Y-Intercept
The y-intercepts of any function are the y-values when x = 0
- Substitute x = 0 into the function:
[tex]f(x)=\frac{x^3-8}{x^2-6x+8}=\frac{0^3-8}{0^2-6(0)+8}}=\frac{-8}{8} =-1[/tex]