{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Stitz-Zeager_College_Algebra_e-book

# 30 centimeters 431 variation in many instances in the

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: To make this exclusion speciﬁc, we write h(x) = (2x+1)(x+1) , x = −1. x+2 3. To ﬁnd the x-intercepts, as usual, we set h(x) = 0 and solve. Solving (2x+1)(x+1) = 0 yields x+2 x = − 1 and x = −1. The latter isn’t in the domain of h, so we exclude it. Our only x2 1 intercept is − 2 , 0 . To ﬁnd the y -intercept, we set x = 0. Since 0 = −1, we can use the reduced formula for h(x) and we get h(0) = 1 for a y -intercept of 0, 1 . 2 2 4. From Theorem 4.1, we know that since x = −2 still poses a threat in the denominator of the reduced function, we have a vertical asymptote there. As for x = −1, we note the factor (x + 1) was canceled from the denominator when we reduced h(x), and so it no longer causes trouble there. This means we get a hole when x = −1. To ﬁnd the y -coordinate of the hole, we substitute x = −1 into (2x+1)(x+1) , per Theorem 4.1 and get 0. Hence, we have a hole on x+2 12 Bet you never thought you’d never see that stuﬀ again before the Final Exam! 4.2 Graphs of Rational Functio...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online