# How do you graph (2x^2+3x+1)/(x^2-5x+4) using asymptotes, intercepts, end behavior?

Dec 28, 2016

Asymptotes $x = 1$, $x = 4$, $y = 2$, intercepts $\left(- \frac{1}{2} , 0\right)$, $\left(- 1 , 0\right)$, $\left(0 , \frac{1}{4}\right)$.

#### Explanation:

Both the numerator and denominator can be factorized, giving
$\frac{\left(2 x + 1\right) \left(x + 1\right)}{\left(x - 1\right) \left(x - 4\right)}$.

Clearly this has two vertical asymptotes $x = 1$ and $x = 4$ (of the simple "$\frac{1}{x}$" type, and two $x$-axis intercepts at $x = \frac{1}{2}$ and $x = - 1$. Putting $x = 0$ gives the $y$-axis intercept as $\left(0 , \frac{1}{4}\right)$,

Since the numerator and denominator have highest powers of $x$ which are the same, namely $2$, there is a horizontal asympote $y = 2$, the $2$ being the ratio of the coefficients of those highest equal powers. (Informally, as $x \to \infty$, the two squares dominate over the other terms in the expression, so all the $x$s nearly cancel out, leaving the constant $2$.)

Going from left to right, the graph must descend steadily from $y = 2$ as $x$ increases from $- \infty$, then go through the two zeroes (and hence through some minimum between the zeroes), thence through the asympototic region at $x = 1$, going through a maximum on its way to the second asymptotic region at $x = 4$, thence decreasing steadily to the horizontal asymptote as $x \to \infty$.