How do you find the domain and range of $(y^{2}) (x - 4) = 8$ $(y^2)(x-4)=8$ ?

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Answer 1 · 2018-01-23T19:10:57

Given: $(y^{2}) (x - 4) = 8$ $(y^2)(x-4)=8$

$y^{2} = \frac{8}{x - 4}$ $y^2 = 8/(x-4)$

$y = \pm \frac{2 \sqrt{2}}{\sqrt{x - 4}}$ $y = +-(2sqrt2)/sqrt(x-4)$

Separate into two equations:

$y = - \frac{2 \sqrt{2}}{\sqrt{x - 4}}$ $y = -(2sqrt2)/sqrt(x-4)$ and $y = \frac{2 \sqrt{2}}{\sqrt{x - 4}}$ $y = (2sqrt2)/sqrt(x-4)$

In both cases, the expression under the radical cannot be negative and it must be greater than 0:

$x - 4 > 0$ $x -4 > 0$

$x > 4 \leftarrow$ $x > 4 larr$ this is the domain for both equations.

Find the range for the first equation:

$lim_(x to 4) -(2sqrt2)/sqrt(x-4) = -oo$

$lim_(x to oo) -(2sqrt2)/sqrt(x-4) = 0$

The range for the first equation is $-oo < y < 0$

$lim_(x to 4) (2sqrt2)/sqrt(x-4) = oo$

$lim_(x to oo) (2sqrt2)/sqrt(x-4) = 0$

The range for the second equation is $0 < y < oo$

How do you find the domain and range of (y^2)(x-4)=8(y2)(x−4)=8(y^2)(x-4)=8?