Question

1. graph the rational function $y = \frac{x - 1}{x - 2}$. both branches of the rational function pass through which quadrant?

\bigcirc quadrant 2
\bigcirc quadrant 3
\bigcirc quadrant 1
\bigcirc quadrant 4

Explanation:

Step1: Simplify the function

The function $y=\frac{x-1}{x-2}$ can be rewritten as $y = 1 + \frac{1}{x-2}$ by polynomial division: $\frac{x-1}{x-2} = \frac{(x-2)+1}{x-2} = 1 + \frac{1}{x-2}$.

Step2: Analyze the transformed reciprocal function

The basic reciprocal function $y=\frac{1}{x}$ passes through quadrants 1 and 3. Shifting it right by 2 units ($y=\frac{1}{x-2}$) and up by 1 unit ($y=1+\frac{1}{x-2}$) does not change the quadrants it covers—both branches still lie in quadrants 1 and 3.

Step3: Verify with test points

• For $x>2$ (right of vertical asymptote $x=2$): let $x=3$, then $y=\frac{3-1}{3-2}=2$ (quadrant 1).
• For $x<2$ (left of vertical asymptote $x=2$): let $x=1$, then $y=\frac{1-1}{1-2}=0$; let $x=0$, then $y=\frac{0-1}{0-2}=0.5$? No, wait: $x=0$, $y=\frac{-1}{-2}=0.5$ (quadrant 1? No, $x=0$ is on y-axis. Let $x=-1$: $y=\frac{-1-1}{-1-2}=\frac{-2}{-3}\approx0.67$ (quadrant 3? No, $x=-1$ (negative), $y=0.67$ (positive) is quadrant 2? Wait correction: $x=-1$, $y=\frac{-2}{-3}=\frac{2}{3}$: $x<0,y>0$ is quadrant 2; $x=3,y=2$ (quadrant1); $x=1.5$ (between 1 and2): $y=\frac{0.5}{-0.5}=-1$ (quadrant4); $x=4,y=\frac{3}{2}=1.5$ (quadrant1); $x=-2,y=\frac{-3}{-4}=0.75$ (quadrant2); $x=0.5,y=\frac{-0.5}{-1.5}=\frac{1}{3}$ (quadrant2); $x=5,y=\frac{4}{3}\approx1.33$ (quadrant1). Wait original mistake: the two branches are: 1. $x>2$: $y>1$ (quadrant1); 2. $x<2$: when $x<2$, $\frac{1}{x-2}<0$, so $y=1+\text{negative}$: if $x-2<-1$ (i.e., $x<1$), $\frac{1}{x-2}\in(-1,0)$, so $y\in(0,1)$ (quadrant2); if $1

Answer:

quadrant 1, quadrant 3