Chapter 4.5, Problem 4AYU

To determine

To find: Make up a rational function that has $y=2x+1$ as an oblique asymptote. Explain the methodology that you used.

Answer to Problem 4AYU

$\frac{P\left(x\right)}{Q\left(x\right)} =\frac{2x^2 + x + 1}{x}$

Explanation of Solution

Recall that to find the oblique asymptote of a rational function, we perform division; the quotient will be the equation of the oblique asymptote. Of course, the oblique asymptote is a line, so we require that the quotient is linear and, consequently, the dividend is a polynomial one degree higher than the divisor.

So, since $y=2x+1$ is the oblique asymptote of some rational function:

$\frac{P\left(x\right)}{Q\left(x\right)} = \left(2x + 1\right)+\frac{R\left(x\right)}{Q\left(x\right)}$, where $R\left(x\right)$ is the remainder

$P\left(x\right) = \left(2x + 1\right)Q\left(x\right) + R\left(x\right)$

Now, $P\left(x\right)$'s degree must be one more than $Q\left(x\right)$’s degree. If we pick anything for $Q\left(x\right)$, say $Q\left(x\right)=x$, then $P\left(x\right) = x\left(2x + 1\right) + R\left(x\right) = 2x^2 + x + R\left(x\right)$.

$P\left(x\right)$ must be a quadratic, so $R\left(x\right)$ must be either a constant or a linear polynomial. If $R\left(x\right) = 1$, then: $P\left(x\right) = 2x^2 + x + 1$.

Thus, one such function is: $\frac{P\left(x\right)}{Q\left(x\right)} =\frac{2x^2 + x + 1}{x}$.