# How do you graph y=(34x-2)/(16x+4) using asymptotes, intercepts, end behavior?

Nov 23, 2016

I have managed to present the rectangular hyperbola represented by your equation.. Vertical asymptote: $x = - \frac{1}{4}$; the horizontal one: $y = \frac{17}{2}$. See explanation, for more.

#### Explanation:

graph{(y-17/2) (x+1/4)+9/4=0 [-40, 40, -20, 20]}

By actual division and reorganization,

$\left(y - \frac{17}{2}\right) \left(x + \frac{1}{4}\right) = - \frac{9}{4}$

This represents athe rectangular hyperbola with asymptotes y

=17/2 and x = -1/4.

The center is at the common point $\left(- \frac{1}{4} , \frac{17}{2}\right)$..

The graph would clarify these statements.

As $y \to \frac{17}{2} , x \to \pm \infty$.

As $x \to - \frac{1}{4} , y \to \pm \infty$.

Not easy to see in the graph, the x-intercept is 1/68 and y-intercept

is $-$1/2.

Note that the general equation of a rectangular hyperbola is of the

form

$\left(y - m x + a\right) \left(x + m y + b\right) = c$,

enabling us to read the equations of the asymptotes as

$y - m x + a = 0 = x + m y + b$