How do you find the domain and range of $y= x^2 / (x^2-16)$ ?

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Answer 1 · 2017-11-17T06:56:24

The domain is $x in (-oo,-4) uu(-4,4) uu (4,+oo)$ . The range is $y in (-oo,1) uu (1,+oo)$

As we cannot divide by $0$ , the denominator is

$x^2-16!=0$

$x^2!=16$

$x!=4$ and $x!=-4$

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

Calculation of the range :

$y=x^2/(x^2-16)$

$y(x^2-16)=x^2$

$yx^2-16y=x^2$

$yx^2-x^2=16y$

$x^2(y-1)=16y$

$x^2=(16y)/(y-1)$

$x=+-sqrt((16y)/(y-1))$

Therefore,

$y!=1$

The range is $y in (-oo,1) uu (1,+oo)$

graph{x^2/(x^2-16) [-10, 10, -5, 5]}

How do you find the domain and range of y= x^2 / (x^2-16)y= x^2 / (x^2-16)?