Lets say we have
P(x)/q(x)

vertical assymtotes are in the form x=something, not y=0
y=0 are horizontal assemtotes

so verticall assymtotes
reduce the fraction
set the denomenator equal to zero
those values that make the deomenator zero are the vertical assymtotes

the horizontal assymtote
when the degree of P(x)<q(x), then HA=0

when the degree of P(x)=q(x), then divide the leading coefient of P(x) by the leading coeficnet of q(x)
example, f(x)=(2x^2-3x+3)/(9x^2-93x+993), then HA is 2/9

ok so for vertical assymtote example
f(x)=x/(x^2+5x+6)
the VA's are at x=-3 and x=-2

horizontal assymtote
make degree same
f(x)=(3x^2-4)/(8x^2+9x),
the HA is 3/8
hope I helped, read the whole thing then ask eusiton

6 0

4 years ago

Evaluate M/N for M=10, N=-5, and P =-2

Elden [556K]

so it would be 10/-5 which would be -2

7 0

4 years ago