What is the domain and range of #f(x) =(x-3)/(7x+4)#?

So this is a rational function, so to get #f(x)# into hyperbolic form (from which you can read off your asymptotes) perform polynomial division.

# x/(7x) = 1/7#

# 1/7*7x = x# and #1/7*4=4/7#

#x - x =0# and #-3 - 4/7 = -3 4/7 = -25/7#.

So your quotient, #Q(x) = 1/7# and your remainder #R(x) = -25/7#.

Therefore, #f(x)=1/7 + -25/7 -: 7x +4#

Rearrange this to make your solution of the form #f(x) = a/(x-h) +k#

Therefore, #f(x) = -25/ (49(x+4/7)) + 1/7#

This means that you have asymptotes at #x=-4/7# and #y=1/7#. Since hyperbolas have infinite domains and ranges excluding any point on the asymptote, you have:

#dom f in RR \\\ {-4/7}# and #ran f in RR \\\ {1/7}#.

#"f(x) is defined for all real values of x apart from values that"#
#"make the denominator equal to zero"#

#"equating the denominator to zero and solving gives the"#
#"value that x cannot be"#

#"solve " 7x+4=0rArrx=-4/7larrcolor(red)" excluded value"#

#rArr"domain is " x inRR,x!=-4/7#

#"to find any excluded value in the range, rearrange "#

#y=f(x)" making x the subject"#

#rArry(7x+4)=x-3larrcolor(blue)" cross-multiplying"#

#rArr7xy+4y=x-3#

#rArr7xy-x=-3-4y#

#rArrx(7y-1)=-(3+4y)#

#rArrx=-(3+4y)/(7y-1)#

#"the denominator cannot equal zero"#

#"solve " 7y-1=0rArry=1/7larrcolor(red)" excluded value"#

#rArr"range is " y inRR,y!=1/7#