Question

Question #911b5

Answer 1

`(-oo , oo)`

Explanation:

`f(x)=(x^2-5x+4)/(x^2-3x+2)`

Vertical asymptotes occur where the function is undefined. In this case where the denominator is `0`

`x^2-3x+2=0`

`(x-1)(x-2)=0=>x=1 and x=2`

If we factor the numerator:

`x^2-5x+4`

`(x-4)(x-1)`

When we view the function like this:

`f(x)=((x-4)(x-1))/((x-1)(x-2))`

We notice that we can cancel the `(x-1)` factors. This is known as removing the discontinuity. So we now know that the vertical line `x=2` is an asymptote, and `x=1` is just a point discontinuity.

as `x->2^-` , `(x^2-5x+4)/(x^2-3x+2)->oo`

as `x->2^+` , `(x^2-5x+4)/(x^2-3x+2)->-oo`

So range of function is:

`(-oo , oo)`

Answer 2

The range is `y in RR-{1}`

Explanation:

We are given

`y=(x^2-5x+4)/(x^2-3x+2)=((x-4)cancel(x-1))/((x-2)cancel(x-1))`

Rearranging the function

`y(x-2)=(x-4)`

`xy-2y=x-4`

`xy-x=2y-4`

`x(y-1)=2(y-2)`

`x=2(y-2)/(y-1)`

`x` will have solutions iff the denominator is `!=0`

Therefore,

`y-1!=0`

`y!=1`

The range is `y in RR-{1}`
graph{(x-4)/(x-2) [-11.25, 11.245, -5.63, 5.62]}