What is the domain and range of $\frac{x^{2} + 2}{x + 4}$ $(x^2+2)/(x+4)$ ?

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Answer 1 · 2018-04-06T19:38:52

The domain is $x in RR-{-4}$ . The range is $y in (-oo, -16.485] uu [0.485, +oo)$

The denominator is $!=0$

$x+4!=0$

$x!=-4$

The domain is $x in RR-{-4}$

To find the range, proceed as follws

Let $y=(x^2+2)/(x+4)$

$y(x+4)=x^2+2$

$x^2-yx+2-4y=0$

This is a quadratic equation in $x^2$ and in order to have solutions

the discriminant $Delta>=0$

Therefore

$Delta=(-y)^2-4(1)(2-4y)>=0$

$y^2-16y-8>=0$

The solutions are

$y=(-16+-sqrt((-16)^2-4(1)(-8)))/2=(-16+-16.97)/2$

$y_1=-16.485$

$y_2=0.485$

The range is $y in (-oo, -16.485] uu [0.485, +oo)$

graph{(x^2+2)/(x+4) [-63.34, 53.7, -30.65, 27.85]}

What is the domain and range of (x^2+2)/(x+4)x2+2x+4(x^2+2)/(x+4)?