How do you graph #f(x)=(2x^2 + 5) / (x-1)#?

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Answer 1 · 2015-08-28T04:09:39

graph{2(x+1)+7/(x-1) [-20, 20, -20, 20]}

I can suggest to transform the function (defined everywhere but a point #x=1# when the denominator equals to zero) into the following form:

#f(x) = (2x^2+5)/(x-1) = (2x^2-2+7)/(x-1) = [2(x-1)(x+1)+7]/(x-1)=#

#= 2(x+1)+7/(x-1)#

In this form we can graph separately
#y = 2(x+1) = 2x+2# and
#y=7/(x-1)#,
after which we just add two graphs.

#y=2x+2#

graph{2x+2 [-10, 10, -5, 5]}

#y=7/(x-1)#

graph{7/(x-1) [-10, 10, -5, 5]}

The sum of these graphs is

#y= 2(x+1)+7/(x-1)#

graph{2(x+1)+7/(x-1) [-20, 20, -20, 20]}

Notice that #x=1# is an asymptote of this graph