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]}

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