Answer:
vertical asymptote at x = 7
horizontal asymptote at y = 6
Step-by-step explanation:
The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.
solve:
x − 7 = 0 ⇒ x = 7 is the asymptote
Horizontal asymptotes occur as
lim ,f(x)→ c(a constant)
x → ± ∞
divide terms on numerator/denominator by x
f ( x ) = 5/ x +6= 5 /x + 6
x/ x − 7 /x 1 − 7/ x
as x ± ∞ , f ( x ) → 0 +6
1 − 0
⇒ y = 6 is the asymptote
graph{((5)/(x-7))+6 [-20, 20, -10, 10]}
[tex]\bf f(x)=\cfrac{5}{x+7}+6\implies f(x)=\cfrac{5+6(x+7)}{x+7}\implies f(x)=\cfrac{6x+47}{1x+7}[/tex]
to get the vertical asymptotes, we simply zero out the denominator and see what's the equation so let's do so
x + 7 = 0............. x = -7....... that's the vertical asymptote.
to get the horizontal asymptotes, there are 3 cases
if the denominator's degree is higher, or lower or the exactly the same as the degree of the numerator.
well, notice the rational, the degree of the numerator is just 1, 6x¹, and the degree of the denominator is also 1, 1x¹.
when that occurs, the horizontal asymptote occurrs at the fraction of their leading term coefficients.
[tex]\bf f(x)=\cfrac{6x+47}{1x+7}~\hspace{10em}\stackrel{\textit{horizontal asymptote}}{y=\cfrac{6}{1}}\implies y=6 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{V~asymptote}{x=-7}\qquad \stackrel{H~asymptote}{y=6}~\hfill[/tex]
