Question

identify the oblique asymptote of f(x) = (x-1)/(4x^2+2x-3)

Answer 1

y = 4x + 6

y = 4x – 2

y = 0

No Oblique Asymptote

Answer 2

there will be no oblique asymptote because the degree of the denominator is greater than the degree of the numerator.

Answer 3

\[f \left( x \right)=\frac{ \left( x-1 \right) }{ 4x ^{2}+2x-3 }\]

like this?

Answer 4

yes!

Answer 5

what @Data_LG2 said

Answer 6

in order to get a slant or oblique asymptote, the degree of the polynomial in the numerator must be 1 more than the degree of the polynomial in the denominator.

Did you give this a read?

http://www.purplemath.com/modules/asymtote3.htm

Answer 7

yeah i read it

Answer 8

get anything from it?

Answer 9

yeah, i kind of understand, its just easier for me to learn through a teacher in person, because it takees me a while to grasp things so idk

Answer 10

i understand... do your best. make sure to point out which parts you're having troubles with so you can learn.

Answer 11

1) degree of bottom larger that degree of the top: horizontal asymptote \(y=0\) aka the \(x\) axis

2) degrees are the same, horizontal asymptote is \(y=\text{ratio of leading coefficients}\)

3) degree of top is one more than degree of bottom: no horizontal asymptote, slant asymptote is what you get when you divide and ignore the remainder

Answer 12

so is it y=0?

Answer 13

because the degree on the bottom is greater than the degree on the top

Answer 14

good job!

Answer 15

yes, but not oblique

Answer 16

thank you!!

Answer 17

yes, it is a horizontal asymptote.