Question

How do you find the domain and range of `(x-2)/(x^2+3x-10)`?

Answer 1

The domain is `x in RR-{-5}`. The range is `y in (-oo,0) uu (0, +oo)`

Explanation:

The denominator is

`x^2+3x-10=(x-2)(x+5)`

Therefore,

`(x-2)/(x^2+3x-10)=(x-2)/((x-2)(x+5))=1/(x+5)`

As the denominator `!=0`, so

`(x+5)!=0`

So,

The domain is `x in RR-{-5}`

To calculate the range, proceed as follows

Let `y=1/(x+5)`

`y(x+5)=1`

`yx+5y=1`

`x=(1-5y)/(y)`

So,

`y!=0`

Therefore,

The range is `y in (-oo,0) uu (0, +oo)`

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

Answer 2

`x inRR,x!=-5`
`y inRR,y!=0`

Explanation:

`"let "y=(x-2)/(x^2+3x-10)`

`"factorise numerator/denominator and simplify"`

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

`"the denominator cannot equal zero as this would make"`
`"y undefined. Equating the denominator to zero and"`
`"solving gives the value that x cannot be"`

`"solve "x+5=0rArrx=-5larrcolor(red)"excluded value"`

`rArr"domain is "x inRR,x!=-5`

`"to find the range rearrange making x the subject"`

`y(x+5)=1larrcolor(blue)"cross-multiply"`

`rArrxy+5y=1`

`rArrxy=1-5y`

`rArrx=(1-5y)/y`

`"the denominator cannot equal zero"`

`rArr"range is "y inRR,y!=0`