<img src="https://tex.z-dn.net/?f=F%28x%29%20%3D%20%5Cfrac%7Bx%5E%7B3-8%7D%20%7D%7Bx%5E%7B2%7D%20-6x%2B8%7D" id="TexFormula1" ti

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3 years ago

tle="F(x) = \frac{x^{3-8} }{x^{2} -6x+8}" alt="F(x) = \frac{x^{3-8} }{x^{2} -6x+8}" align="absmiddle" class="latex-formula">
What is the
Domain:
Holes:
VA:
HA:
OH:
Roots:
Y-Intercept

Mathematics

1 answer:

Firlakuza [10]3 years ago

3 0

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:

$f(x)=\frac{x^3-8}{x^2-6x+8}$

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 → ∞

$\lim_{x \to \infty}(x + 6) + \frac{28x-56}{x^2-6x+8}=x+6$

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:

$f(x)=\frac{x^3-8}{x^2-6x+8}=\frac{0^3-8}{0^2-6(0)+8}}=\frac{-8}{8} =-1$

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We need to conduct a hypothesis in order to check if the mean is more than 3.1, the system of hypothesis would be:

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We can replace in formula (1) the info given like this:

$t=\frac{3.25-3.1}{\frac{0.3}{\sqrt{36}}}=3$

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This rule is illustrated in the attached image.

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