# How do you graph y=(3x+5)/(2x-11) using asymptotes, intercepts, end behavior?

Dec 9, 2017

See below

#### Explanation:

$f \left(x\right) = \frac{3 x + 5}{2 x - 11}$

$D \left(f\right) = \left(- \infty , \frac{11}{2}\right) \bigcup \left(\frac{11}{2} , \infty\right)$
Since this function isn't defined in $x = \frac{11}{2}$ It has an asymptote there.

Y intercept: only when x=0
$y = \frac{3 \cdot 0 + 5}{2 \cdot 0 - 11} = - \frac{5}{11}$
$\left[0 , - \frac{5}{11}\right]$

X intercept> onlz when y=0
$0 = \frac{3 x + 5}{2 x - 11}$
$0 \cdot 2 x - 11 = \frac{3 x + 5}{2 x - 11} \cdot 2 x - 11$
$0 = 3 x + 5$
$- \frac{5}{3} = x$
$\left[- \frac{5}{3} , 0\right]$

End bahaviour in $+ \infty$
$L i {m}_{x \rightarrow \infty} \frac{3 x + 5}{2 x - 11} \cdot \frac{\frac{1}{x}}{\frac{1}{x}} = L i {m}_{x \rightarrow \infty} \frac{3 + \frac{5}{x}}{2 - \frac{11}{x}} = \frac{3 + 0}{2 - 0} = \frac{3}{2}$

End bahaviour in $- \infty$(should be the same)
$L i {m}_{x \rightarrow - \infty} \frac{3 x + 5}{2 x - 11} \cdot \frac{\frac{1}{x}}{\frac{1}{x}} = L i {m}_{x \rightarrow - \infty} \frac{3 + \frac{5}{x}}{2 - \frac{11}{x}} = \frac{3 - 0}{2 + 0} = \frac{3}{2}$

so another asymptote is: $y = \frac{3}{2}$

Since we know it's Reciprocal Function. There's another way to graph this function which is to rewrite f(x) to this formula:
$y = \frac{k}{x - a} + b$

$f \left(x\right) = \left[\frac{\left(3 x + 5\right)}{\left(2 x - 11\right)} \cdot \frac{2}{3}\right] \cdot \frac{3}{2} = \left[\frac{6 x + 10}{6 x - 33}\right] \cdot \frac{3}{2} = \left[\frac{6 x - 33 + 33 + 10}{6 x - 33}\right] \cdot \frac{3}{2} = \left[\frac{6 x - 33}{6 x - 33} + \frac{43}{6 x - 33}\right] \cdot \frac{3}{2} = \left[1 + \frac{43}{3 \left(2 x - 11\right)}\right] \cdot \frac{3}{2} = \frac{3}{2} + \frac{43 \cdot \cancel{3}}{\cancel{3} \cdot 2 \left(2 x - 11\right)} = \frac{3}{2} + \frac{43}{4 x - 22}$
$y = \frac{43}{4 x - 22} + \frac{3}{2}$

$k = 43 > 0$ This means it is in 1. and 3. quadrant
1. asymptote: $y = \frac{3}{2}$
2. asymptote: $x = \frac{22}{4} = \frac{11}{2}$