How do you graph f(x)=2/x^2+1 using holes, vertical and horizontal asymptotes, x and y intercepts?

zakownikbj

zakownikbj

Answered question

2022-09-21

How do you graph f ( x ) = 2 x 2 + 1 using holes, vertical and horizontal asymptotes, x and y intercepts?

Answer & Explanation

Trace Arias

Trace Arias

Beginner2022-09-22Added 6 answers

f ( x ) = 2 x 2 + 1
Since x 2 + 1 > 0 x there exists no holes in f(x)
Also, lim x-> +-oo f ( x ) = 0
f ( x ) = - 4 x ( x 2 + 1 ) 2
For a maximum or minimum value; f'(x)=0
- 4 x ( x 2 + 1 ) 2 = 0 x = 0
f ( 0 ) = 2 0 + 1 = 2
Since f ( 0 ) = 2 is a maximum of f(x)
The critical points of f(x) can be seen on the graph below:
graph{2/(x^2+1) [-5.55, 5.55, -2.772, 2.778]}

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?