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

2 Answers
Nov 25, 2017

Answer:

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]}

Nov 25, 2017

Answer:

#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#