Determine the equations of the asymptotes of the graph for the function f(x)=x/5x+2

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Determine the equations of the asymptotes of the graph for the function f(x)=x/5x+2

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

va = -2/5 ha = 1/5
assuming that 5x+2 is the denominator....
yes it is , thank you ! could you show me how you achieved that. Im clueless.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

va..vertical asymptote...is whatever turns the bottom to zero; 5x+2 =0 5x = -2 x = -2/5 is the va
the ha is found by dividing top and bottom by the highest power of x in the bottom; in this case that is just 'x' x/x 1 -------- = ------- 5x/x + 2/x 5 + (2/x)
with it like this; we see that when x get very large; 1 ------------ = .0000...00001 - is very tiny 10000...0000
so 2/x gets close to zero and fades away... in fact, anything with an x left on bottom goes to zero and we are left with: 1/5
ohh right , so to find the the ha you always have to take the variable x into consideration
yes
The Rule for horizontals is that it depends on the degree of the top as compared to the degree of the bottom. If the degree of the top is smaller, then you have a horizontal asymptote on the x-axis. If the degrees are the same, then you have a horizontal asymptote on the line y=a/b where a is the leading coefficient on the top and b is the leading coefficient on the bottom. If the degree of the bottom is smaller, then you don't have a horizontal asymptote - you have an oblique one instead. I teach my students to find those using synthetic division and ignoring the remainder but you can just use regular polynomial long-division
and if you want to read a novel..there it is :)
thank you both for your help !
:) youre welcome.
sorry for butting in - someone told me about this site and I was looking and saw this question. I am a teacher and gave a final exam that had this question on it yesterday so thought I'd try to help. Cool site - I'm going to tell my students about it!

Not the answer you are looking for?

Search for more explanations.

Ask your own question