How do you find the range of a rational function algebraically $y = \frac{- 4 x - 3}{x - 2}$ $y=(-4x-3)/(x-2)$ ?

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Answer 1 · 2018-04-09T09:26:50

the range is all reals except when $y = - 4$ $y=-4$

$y = \frac{- 4 x - 3}{x - 2}$ $y=(-4x-3)/(x-2)$

$y = \frac{- 4 (x - 2) - 11}{x - 2}$ $y=(-4(x-2)-11)/(x-2)$

$y = - 4 - \frac{11}{x - 2}$ $y=-4-11/(x-2)$

Therefore, the range is all reals except when $y = - 4$ $y=-4$

How do you find the range of a rational function algebraically y=−4x−3x−2y=(-4x-3)/(x-2)?