Step-by-step explanation:
We have given,
A rational function : f(x) = [tex]\frac{x-2}{x-4}[/tex]
W need to find :
Point of discontinuity : - At x = 4, f(x) tends to reach infinity, So we get discontinuity point at x =4.
For no values of x, we get indetermined form (i.e [tex]\frac{0}{0}[/tex]), Hence there is no holes
Vertical Asymptotes:
Plug y=f(x) = ∞ in f(x) to get vertical asymptote {We can us writing ∞ = [tex]\frac{1}{0}[/tex]}
i.e ∞ = [tex]\frac{x-2}{x-4}[/tex]
or [tex]\frac{1}{0}=\frac{x-2}{x-4}[/tex]
or x-4 =0
or x=4, Hence at x = 4, f(x) has a vertical asymptote
X -intercept :
Plug f(x)=0 , to get x intercept.
i.e 0 = [tex]\frac{x-2}{x-4}[/tex]
or x - 2 =0
or x = 2
Hence at x=2, f(x) has an x intercept
Horizontal asymptote:
Plug x = ∞ in f(x) to get horizontal asymptote.
i.e f(x) = [tex]\frac{x-2}{x-4}[/tex] = [tex]\frac{x(1-\frac{2}{x} )}{x(1-\frac{4}{x} )}[/tex]
or f(x) = [tex]\frac{(1-\frac{2}{∞} )}{(1-\frac{4}{∞} )}[/tex]
or f(x) = 1 = y
hence at y =f(x) = 1, we get horizontal asymptote
