Rational Functions Comprehensive II

Problem 273: For the rational function

$y = \frac{7(x-1)}{(x-4)(x-7)},$

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.

$(x-4)(x-7)= 0 \implies x - 4 = 0 \ \text{or} \ x - 7 = 0 \implies x = 4, 7$

Thus the domain is

$(-\infty, 4) \cup (4, 7) \cup (7, \infty).$

(b) Since the degree of the polynomial in the denominator is bigger than the degree of the polynomial in the denominator, the horizontal asymptote is given by the line $y = 0$ .

That is, the horizontal asymptote is $y =0$ .

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

$(x - 4)(x - 7) = 0 \implies x - 4 = 0 \ \text{or} \ x - 7 = 0 \implies x = 4, 7$

That is, the vertical asymptotes are $x = 4, 7$ .

(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.

$0 = \frac{7(x-1)}{(x - 4)(x - 7)} \implies 7(x - 1) = 0 \implies x = 1 \implies (1, 0)$

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

$y = \frac{7(0-1)}{(0-4)(0-7)} = \frac{7(-1)}{(-40(-7)} = \frac{-7}{28} \implies (0, -1/4).$

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