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

Question

1904 views around the world

You can reuse this answer
Creative Commons License

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

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