Find-the-range-and-domain-of-y-x-1-x-2-3x-10-x-2-6x-8-

Question Number 5235 by sanusihammed last updated on 02/May/16

F i n d t h e r a n g e a n d d o m a i n o f

y = \frac{(x + 1) (x^{2} + 3 x - 10)}{x^{2} + 6 x + 8}

Answered by Yozzii last updated on 02/May/16

x^{2} + 6 x + 8 = {(x + 3)}^{2} - 1 .

I f x^{2} + 6 x + 8 = 0

\Rightarrow {(x + 3)}^{2} - 1 = 0

\Rightarrow x + 3 = \pm 1

x = - 3 \pm 1 \Rightarrow x = - 2, - 4 .

\frac{x^{2} + 6 x + 8 - 3 x - 18}{x^{2} + 6 x + 8} = 1 - \frac{3 x + 18}{x^{2} + 6 x + 8}

\Rightarrow y = (x + 1) (1 - \frac{3 x + 18}{x^{2} + 6 x + 8})

y = x + 1 - \frac{(x + 1) (3 x + 18)}{x^{2} + 6 x + 8}

(x + 1) (3 x + 18) = 3 x^{2} + 21 x + 18

\frac{3 x^{2} + 21 x + 18}{x^{2} + 6 x + 8} = \frac{3 (x^{2} + 6 x + 8 - 2 + x)}{x^{2} + 6 x + 8}

= 3 + \frac{3 x - 6}{x^{2} + 6 x + 8}

∴ y = x - 2 - \frac{3 x - 6}{x^{2} + 6 x + 8}

y = x - 2 - \frac{3 - \frac{6}{x}}{x + 6 + \frac{8}{x}}

\Rightarrow \lim_{x \to \infty} y = \lim_{x \to \infty} (x - 2 - \frac{3 - 6 x^{- 1}}{x + 6 + \frac{8}{x}}) = \infty - 2 - \frac{3 - \frac{6}{\infty}}{\infty + 6 + \frac{8}{\infty}} = \infty - \frac{3 - 0}{\infty + 9} = \infty (l i m i t d o e s n o t e x i s t)

A l s o, \lim_{x \to - \infty} y = - \infty .

I t i s s a i d t h a t y = x - 2 i s a n o b l i q u e

a s y m p t o t e t o t h e c u r v e y = f (x) .

x = - 2 a n d x = - 4 a r e v e r t i c a l a s y m p t o t e s .

∴ d o m a i n (y) = D_{y} = {x \in R ∣ x \neq - 2, - 4}

r a n g e (y) = I m (y) = R .

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