Factor out the coefficient of the variable for the expression 1/4y + 3/8

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Firlakuza [10]

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.