Rational Functions Comprehensive

Problem 272: For the rational function

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

find the domain, horizontal asymptote, vertical asymptote, holes, $x$ -intercept, and $y$ -intercept.

Solution:

(a) To find the domain, we need to find when the denominator is zero.

$2(x + 4) = 0 \implies x+ 4 = 0 \implies x = -4$

Thus the domain is

$(-\infty, -4) \cup (-4, \infty).$

(b) Since the degree of the polynomial in the denominator and the denominator is the same, the horizontal asymptote is given by the coefficients in front the denominator and numerator.

That is, the horizontal asymptote is $y = -\frac{3}{2}$ .

(c) The vertical asymptote is given by the value of $x$ that would make the denominator equal to 0.

$2(x + 4) = 0 \implies x + 4 = 0 \implies x = -4$

That is, the vertical asymptote is $x = -4$ .

(d) There are not holes in this rational expression since there are not factors to simplify.

(e) Let $y = 0$ to find the $x$ -intercept.

$0 = \frac{-3(x-4)}{2(x + 4)} \implies -3(x - 4) = 0 \implies x = 4 \implies (4, 0)$

(f) Let $x = 0$ to find the $y$ -intercept.

$y = \frac{-3(0-4)}{2(0 + 4)} = \frac{-3(-4)}{2(4)} = \frac{3}{2} \implies (0, 3/2).$

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