fractional polynomial limits

GrahamB08

New member
Joined
Jul 1, 2014
Messages
2
I have a function f(x) = 2x^2 + x - 6/ x^2 +3x+2 for which I am trying to draw a graph.
in this form the value -2 is undefined.

if I factor the function I get (2x-3)(x+2)/(x+1)(x+2)
am I allowed to cancel out the (x+2)?
In the simplified form the value of -2 is defined. I am wondering if I am allowed to just used the simplified form or if I should still consider -2 as undefined.
 
I have a function f(x) = 2x^2 + x - 6/ x^2 +3x+2 for which I am trying to draw a graph.
in this form the value -2 is undefined.

if I factor the function I get (2x-3)(x+2)/(x+1)(x+2)
am I allowed to cancel out the (x+2)?
In the simplified form the value of -2 is defined. I am wondering if I am allowed to just used the simplified form or if I should still consider -2 as undefined.
you should consider -2 as undefined.
you draw graph based on (2x-3)/(x+1), but put a hole in x=-2
 
I have a function f(x) = (2x^2 + x - 6)/ (x^2 +3x+2) for which I am trying to draw a graph.
in this form the value -2 is undefined.

if I factor the function I get (2x-3)(x+2)/(x+1)(x+2)
am I allowed to cancel out the (x+2)?
In the simplified form the value of -2 is defined. I am wondering if I am allowed to just used the simplified form or if I should still consider -2 as undefined.

Those () are important - super important!!!
 
Grahamb08 said:
I have a function f(x) = (2x^2 + x - 6)/ (x^2 +3x+2) for which I am trying to draw a graph.
in this form the value -2 is undefined.

if I factor the function I get (2x-3)(x+2)/((x+1)(x+2)) \(\displaystyle \ \ \ \) <-------- And so are these.
am I allowed to cancel out the (x+2)?
In the simplified form the value of -2 is defined. I am wondering if I am allowed to just used the
simplified form or if I should still consider -2 as undefined.

Subhotosh Khan said:
Those () are important - super important!!!
.
 
Top