How do you find the vertical, horizontal or slant asymptotes for #f(x)= (4x+8)/(x-3)#?

2 Answers
Bdub
Mar 3, 2016

#VA: x-3=0->x=3,HA:f(x)=lim x->prop(4x)/x=4#

Explanation:

Take the denominator set it equal to 0 and solve for x to find the vertical asymptote. To find the horizontal asymptote find the limit as x goes to infinity and use the end behavior of a rational function. That is, pick the highest degree term from the top and bottom then simplify

Leland Adriano Alejandro
Mar 3, 2016

#x=3# is a Vertical Asymptote
#y=4# is a Horizontal Asymptote
Slant Asymptote: None

Explanation:

At #x=3# the value of the function is not defined. Therefore the line approached by the graph but will never touch it no matter how far is #x=3#

Similarly, no matter how far the graph of the function increases or decreases to the right or to the left respectively, the line approached by the graph is #y=4#

Therefore #y=4# is a horizontal asymptote

See the graph of #y= (4x+8)/(x-3)#

graph{(y- (4x+8)/(x-3))=0[-40,40,-20,20]}

See also the graph of the asymptotes #x=3# and #y=4#

graph{(y-4)(y+(1000)x-3*(1000))=0[-40,40,-20,20]}

God bless ... I hope the explanation is useful

Related questions
See all questions in Asymptotes
Impact of this question
1433 views around the world
You can reuse this answer
Creative Commons License