Answer:
The discontinuity is x = 4
There no holes
The equation of the vertical asymptote is x = 4
The x intercept is 2
The equation of the horizontal asymptote is y = 1
Step-by-step explanation:
* Lets explain the problem
∵ [tex]f(x)=\frac{x-2}{x-4}[/tex]
- To find the point of discontinuity put the denominator = 0 and find
the value of x
∵ The denominator is x - 4
∵ x - 4 = 0 ⇒ add 4 to both sides
∴ x = 4
* The discontinuity is x = 4
- A hole occurs when a number is both a zero of the numerator
and denominator
∵ The numerator is x - 2
∵ x - 2 = 0 ⇒ add 2 to both sides
∴ x = 2
∵ The denominator is x - 4
∵ x - 4 = 0 ⇒ add 4 to both sides
∴ x = 4
∵ There is no common number makes the numerator and denominator
equal to 0
∴ There no holes
- Vertical asymptotes are vertical lines which correspond to the zeroes
of the denominator of the function
∵ The zero of the denominator is x = 4
∴ The equation of the vertical asymptote is x = 4
- x- intercept is the values of x which make f(x) = 0, means the
intersection points between the graph and the x-axis
∵ f(x) = 0
∴ [tex]\frac{x-2}{x-4}=0[/tex] ⇒ by using cross multiplication
∴ x - 2 = 0 ⇒ add 2 to both sides
∴ x = 2
* The x intercept is 2
- If the highest power of the numerator = the highest power of the
denominator, then the equation of the horizontal asymptote is
y = The leading coeff. of numerator/leading coeff. of denominator
∵ The numerator is x - 2
∵ The denominator is x - 4
∵ The leading coefficient of the numerator is 1
∵ The leading coefficient of the denominator is 1
∴ y = 1/1 = 1
* The equation of the horizontal asymptote is y = 1
