Stitz-Zeager_College_Algebra_e-book

The result is a fraction whose numerator is positive

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: cancellation possible, and we conclude that the lines x = − 3 and x = 3 are vertical asymptotes to the graph of y = f (x). The calculator verifies this claim. − x+2 x 2. Solving x2 − 9 = 0 gives x = ±3. In lowest terms g (x) = x x−−−6 = (x−3)(x+2) = x+3 . Since 29 (x 3)(x+3) x = −3 continues to make trouble in the denominator, we know the line x = −3 is a vertical asymptote of the graph of y = g (x). Since x = 3 no longer produces a 0 in the denominator, we have a hole at x = 3. To find the y -coordinate of the hole, we substitute x = 3 into x+2 x+3 2 4.1 Introduction to Rational Functions 237 5 and find the hole is at 3, 6 . When we graph y = g (x) using a calculator, we clearly see the vertical asymptote at x = −3, but everything seems calm near x = 3. The graph of y = f (x) The graph of y = g (x) 3. The domain of h is all real numbers, since x2 + 9 = 0 has no real solutions. Accordingly, the graph of y = h(x) is devoid of both vertical asymptotes and holes. 4. Setting x2 + 4x + 4 = 0 gives us x = −2 as the only real number of concern. Simplifying, x2 − we see r(x) = x2 +4x−6 = (x−3)(x+2) = x−3...
View Full Document

Ask a homework question - tutors are online