let f(x)=(x+1)(((x+1)(x−3)^2 )/((x−1)^2 (x−4) )) then a. find x and y intercepts b. find vertical asymptote and horizontal asymtote c. find domain and range of f d. draw the graph of f


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Question Number 78667 by berket last updated on 19/Jan/20

$l e t f (x) = (x + 1) \frac{(x + 1) {(x - 3)}^{2}}{{(x - 1)}^{2} (x - 4)} t h e n$ $a . f i n d x a n d y i n t e r c e p t s$ $b . f i n d v e r t i c a l a s y m p t o t e a n d h o r i z o n t a l a s y m t o t e$ $c . f i n d d o m a i n a n d r a n g e o f f$ $d . d r a w t h e g r a p h o f f$
	Answered by Rio Michael last updated on 19/Jan/20

	$a . l e t y = \frac{{(x + 1)}^{2} {(x - 3)}^{2}}{{(x - 1)}^{2} (x - 4)}$ $y - i n t e r c e p t, x = 0$ $y = \frac{{(0 + 1)}^{2} {(0 - 3)}^{2}}{{(0 - 1)}^{2} {(0 - 4)}^{2}} = \frac{9}{16}$ $p o i n t (0, \frac{9}{16})$ $x - i n t e r c e p t, y = 0$ $0 = \frac{{(x + 1)}^{2} {(x - 3)}^{2}}{{(x - 1)}^{2} {(x - 4)}^{2}}$ ${[(x + 1) (x - 3)]}^{2} = 0$ $x = - 1 o r x = 3$ $p o i n t (- 1, 0) a n d (3, 0)$ $b . V e r t i c a l a s y m p t o t e s {(x - 1)}^{2} (x - 4) = 0$ $e i t h e r {(x - 1)}^{2} = 0 \Rightarrow x = 1 \lor x = 4$ $H o r i z o n t a l a s y m p t o t e$ $\lim_{x \to \infty} f (x) = \infty h e n c e n o h o r i z o n t a l a s y m p t o t e .$ $c . d o m a i n = v e r t i c a l a s y m p o t e s$
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