Question

y=(x+6)(x+4)/(x-1)^2

Asymptotes, intercepts and domain

Answer 1

I got Domain is all real numbers except 1. Vertical Asymptote = 1. Y int = 24. x int = -6 and -4. Horizontal Asymptote = 0 (not sure on this one)

Answer 2

the HA isn't 0

Answer 3

the degree of the numerator is 2 (when you expand everything out)

Answer 4

the degree of the denominator is also 2

Answer 5

the degrees are equal, so

HA = a/b

where 'a' is the leading coefficient of the numerator and b is the leading coefficient of the denominator

Answer 6

so to do HA... i need to put them into quadratic equations?

Answer 7

so \[x ^{2}+10x+24/x ^{2}-2x+1\]

Answer 8

so in this case the HA is 1?

Answer 9

you got it

Answer 10

no need to actually expand everything out though, you just need the first terms for the numerator and denominator

Answer 11

since 1 is the coefficient in the numerator and the denominator.. right? and am i right then by saying that if I were to do all the math I would pul the x's out and put the constants over x, thus considering x to be very large and those numbers 0? thus why 1/1 ?

Answer 12

yes pretty much, as x gets very large (in either direction), then the x^2 terms really dictate things compared to x terms or the constants since x^2 gets significantly larger than the other terms.

Answer 13

one more general question . in cases where it is (x+b)(x+b).... if x has a coefficient of 2.. would ...so like (2x+b)(3x+a)/(x-c)^2 ... would the top be 6 then? i know.. bad question.. but

Answer 14

In the numerator, the first term is 6x^2

In the denominator, the first term is x^2

The ratio of these terms or their coefficients is 6, so the HA in this case is y = 6

Answer 15

ok.. so u do multiply through ... awesome.. as always.. thanks Jim.. u are appreicated.

Answer 16

yw