Asked by Meron
How can i draw the graph of f(x)=2x^2-3x+2/x^2+1 please help me
Answers
Answered by
Reiny
hard to tell if you mean
f(x)=2x^2-3x+2/x^2+1
or
f(x)=2x^2-3x+2/(x^2+1)
or
f(x)=(2x^2-3x+2)/x^2+1
or
f(x)=(2x^2-3x+2)/(x^2+1)
in any case, make a table of values using reasonable values of x to find the corresponding y values.
Plot your points, join them with a smooth curve
You can test your graph here
http://rechneronline.de/function-graphs/
Make sure you only enter the right side of your equation.
f(x)=2x^2-3x+2/x^2+1
or
f(x)=2x^2-3x+2/(x^2+1)
or
f(x)=(2x^2-3x+2)/x^2+1
or
f(x)=(2x^2-3x+2)/(x^2+1)
in any case, make a table of values using reasonable values of x to find the corresponding y values.
Plot your points, join them with a smooth curve
You can test your graph here
http://rechneronline.de/function-graphs/
Make sure you only enter the right side of your equation.
Answered by
Meron
What is the horizontal asymptote? I think it doesn't have vertical and oblique asymptote. Please please help me please
Answered by
Reiny
Did you not use the webpage I suggested?
You would be able to answer your last questions by looking at the graph.
You also did not answer my questions as to what the function really is
I will assume the last version I posted.
the horizontal asymptote is y = 2
You would be able to answer your last questions by looking at the graph.
You also did not answer my questions as to what the function really is
I will assume the last version I posted.
the horizontal asymptote is y = 2
Answered by
Meron
The webpage gragh and the graph that our teacher draw is not the same what shall i do?
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