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

zakownikbj 2022-09-21 Answered
How do you graph f ( x ) = 2 x 2 + 1 using holes, vertical and horizontal asymptotes, x and y intercepts?
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Answers (1)

Trace Arias
Answered 2022-09-22 Author has 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]}
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