Question

1 8t(2t - 1)(t + 5) Given p(t) answer the following (t + 5)(4 – t)(7t + 2) questions: Round all answers to 3 decimal places as needed. If the answer doesn't exist, type DNE. a) The domain of p in interval notation is: Interval Notation: b) The t-intercept(s) are: c) The vertical asymptotes of p are at t = Write the answer(s) as just numbers separated by commas. d) p has a hole at the ordered pair: Write each hole as an ordered pair. Do not do any rounding until the final answer. Horizontal Asymptote: Write the horizontal asymptote as y = # if it exists. If it doesn't exist, type DNE.

Answer 1

P(t) = 8+(2+-1) (+5) (x+5) (4-+) (76+2) 8+ (2-1) (4-2) (7t+2) Go Domain, denominatus to (t + 5) (4- +) ( 77 +2) #o ht x- * * t# -5 te (-0,-5) u ( -5, - }) u (-3,4 0(4,-)

by t intercept, put plat)=0 87 (24-1) (: (t+5) will get cancelled) t=0 t = = vertical Asymptate denominata SO put (4-6) (7t+2) =0 -2 and t= t = 4 t+5=0 Hole at ds t = -5 p(-5) = 81-5) (26-5) -1) (4-(-5)) (7(-s)+2) 8(-5) (-11) (9) (-33) -1.481 :: (-5, -1.481)

Horizontal Asymptate Degree of Numerator Degree of genommations y = Leading coefficient of numerator Leading coefficient of Denominato (8) (2) (1) 1) (0) (-1)(7) 2.285 no