Explain the degree of the denominator must be exactly one more than that of the numerator for a rational function to have a linear oblique asymptote? Explain

Answer 1

If the numerator p(x) is of degree n, and the denominator q(x) is of degree n-1, then the quotient will be a linear function, with a remainder that approaches zero as x gets large.
For example, (x^3-9x^2+3x-5)/(x^2-x+3) = x-8 + (19-8x)/(x^2-x+3)
Since the remainder has degree one less in the top than in the bottom, it will go to zero as x gets huge, so the rational function p(x)/q(x) is the line y = 19-8x, plus a vanishing remainder.

That is, the slant asymptote is y = 19-8x

Answer 2

let's look at an example.

y = (x^3 + 2x^2 + x + 5) / (x^2 + x + 1)

by long division:

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

now as x ---->∞, the value of (4-x)/(x^2 + x+1) ----> 0
so we are left with y = x+1, which is the linear oblique asymptote

This clearly will happen whenever the degree of the numerator is one more than the degree of the denominator giving us a result of
the form
y = mx+b + R(x)/D(x), where R(x) will be the remainder after the division
and D(x) is the original denominator

Notice the numerator is 1 degree higher than the denominator, we want that
to happen to get a linear result, plus a fraction

In numeric long division, the remainder must always be less than the divisor or else you should increase the quotient, the same is true in algebraic
division, the remainder must have a degree lower than the denominator.
e.g. 456 ÷ 11 = 41 + 5/11 , note 5 < 11
not 456 ÷ 11 = 40 + 16/11, even though this is not incorrect, we should have taken out another 11 to get 41 , leave a remainder of 5

Hope this makes sense to you.