What is the domain and range of #y= (4x) / (x^2 + x - 12) #?

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Answer 1 · 2018-06-09T12:15:15

The domain is #x in (-oo,-4)uu(-4,3)uu(3,+oo)#. The range is #y in RR#

The denominator must be #!=0#

Therefore,

#x^2+x-12!=0#

#(x+4)(x-3)!=0#

#x!=-4# and #x!=3#

The domain is #x in (-oo,-4)uu(-4,3)uu(3,+oo)#

To find the range, proceed as follows

#y=(4x)/(x^2+x-12)#

#=>#, #y(x^2+x-12)=4x#

#=>#, #yx^2+yx-4x-12y=0#

In order for this equation to have solutions, the discriminant #>=0#

Therefore,

#Delta=(y-4)^2-4y*(-12y)#

#=y^2+16-8y+48y^2#

#=49y^2-8y+16#

#AA y in RR, (49y^2-8y+16)>=0#

as #delta=(-8)^2-4*49*16>0#

The range is #y in RR#

graph{(4x)/(x^2+x-12) [-25.66, 25.65, -12.83, 12.84]}