Question

which is the graph of $f(x) = \frac{x - 1}{x^2 - x - 6}$?

Explanation:

Step1: Factor the denominator

Factor \(x^2 - x - 6\): \(x^2 - x - 6=(x - 3)(x + 2)\). So the vertical asymptotes are at \(x = 3\) and \(x=-2\).

Step2: Analyze the x - intercept

Set \(f(x)=0\), \(\frac{x - 1}{(x - 3)(x + 2)}=0\) when \(x - 1 = 0\), so \(x = 1\) is the x - intercept.

Step3: Analyze the behavior of the graph

- For \(x>3\), as \(x\) increases, the numerator \(x - 1\) and denominator \((x - 3)(x + 2)\) are positive, so \(f(x)>0\) and as \(x ightarrow\infty\), \(f(x)\approx\frac{x}{x^2}=\frac{1}{x} ightarrow0^+\).
- For \(2<x<3\), the denominator \((x - 3)(x + 2)<0\) and numerator \(x - 1>0\) (when \(x>1\)) or \(x - 1<0\) (when \(x<1\)). In the interval \(2<x<3\), when \(x\in(2,3)\), \(x - 1>0\) (since \(x>2>1\)), denominator is negative, so \(f(x)<0\).
- For \(-2<x<1\), numerator \(x - 1<0\), denominator \((x - 3)(x + 2)<0\) (because \(x+2>0\) and \(x - 3<0\)), so \(f(x)>0\).
- For \(x<-2\), numerator \(x - 1<0\), denominator \((x - 3)(x + 2)>0\) (because \(x - 3<0\) and \(x + 2<0\)), so \(f(x)<0\).

Now, let's check the graphs:
- The first graph: Check the x - intercept (\(x = 1\)) and the behavior around asymptotes. The graph has a part where for \(x>3\) it goes to positive infinity (wrong, should approach 0 from positive side) and the part between \(2\) and \(3\) has wrong sign.
- The second graph: Check the x - intercept (\(x = 1\)) and the behavior around asymptotes. The graph has a part where for \(x>3\) it goes to positive infinity (wrong, should approach 0 from positive side) and the part between \(2\) and \(3\) has wrong sign.
- The third graph: Wait, no, let's re - evaluate. Wait, the first graph (left - most) has vertical asymptotes at \(x=-2\) and \(x = 3\). Let's check the x - intercept at \(x = 1\). The left - most graph: when \(x>3\), the graph is going up (but should approach 0 from positive side), no. Wait, maybe I made a mistake. Wait, the middle graph: Let's check the x - intercept. The middle graph crosses the x - axis at \(x = 1\)? Wait, no, the first graph (left) has a curve on the left (for \(x<-2\)) going down (since \(f(x)<0\) for \(x<-2\)), a curve between \(-2\) and \(3\) with a peak (since \(f(x)>0\) between \(-2\) and \(1\) and \(f(x)<0\) between \(1\) and \(3\)), and a curve for \(x>3\) going up? No, wait, the correct graph should have:

Wait, let's re - analyze the function \(f(x)=\frac{x - 1}{(x - 3)(x + 2)}\).

Vertical asymptotes at \(x=-2\) and \(x = 3\).

x - intercept at \(x = 1\).

For \(x>3\): \(f(x)=\frac{x - 1}{(x - 3)(x + 2)}\), as \(x ightarrow3^+\), denominator \( ightarrow0^+\), numerator \( ightarrow2\), so \(f(x) ightarrow+\infty\). As \(x ightarrow\infty\), \(f(x)\approx\frac{x}{x^2}=\frac{1}{x} ightarrow0^+\), so the right - hand side of \(x = 3\) should have a curve that comes from \(+\infty\) and approaches \(0\) from above.

For \(2<x<3\) (wait, no, \(x=-2\) and \(x = 3\) are asymptotes, so intervals are \((-\infty,-2)\), \((-2,3)\), \((3,\infty)\)).

In \((-\infty,-2)\): \(f(x)=\frac{x - 1}{(x - 3)(x + 2)}\), numerator \(x - 1<0\), denominator \((x - 3)(x + 2)>0\) (since \(x-3<0\) and \(x + 2<0\)), so \(f(x)<0\). So the graph for \(x<-2\) should be below the x - axis.

In \((-2,3)\): denominator \((x - 3)(x + 2)<0\) (since \(x + 2>0\) and \(x - 3<0\)). Numerator \(x - 1\): when \(x\in(-2,1)\), \(x - 1<0\), so \(f(x)=\frac{\text{negative}}{\text{negative}}=\text{positive}\); when \(x\in(1,3)\), \(x - 1>0\), so \(f(x)=\frac{\text{positive}}{\text{negative}}=\text{negative}\).

In \((3,\infty)\): denominator \((x - 3)(x + 2)>0\), numerator \(x - 1>0\), so \(f(x)>0\), and as \(x ightarrow\infty\), \(f(x) ightarrow0\).

Now, let's check the three graphs:

- First graph (left): For \(x<-2\), the curve is below the x - axis (good, \(f(x)<0\)). For \(-2<x<1\), the curve is above the x - axis (good, \(f(x)>0\)). For \(1<x<3\), the curve is below the x - axis (good, \(f(x)<0\)). For \(x>3\), the curve is above the x - axis and approaches 0 (good, as \(x ightarrow\infty\), \(f(x) ightarrow0^+\)). Also, the x - intercept at \(x = 1\) (the graph crosses the x - axis at \(x = 1\)).

- Second graph (middle): For \(x>3\), the curve is going down (but \(f(x)>0\) for \(x>3\), so it should be above the x - axis, this is wrong).

- Third graph (right): It has only one vertical asymptote (but we have two vertical asymptotes \(x=-2\) and \(x = 3\)), so this is wrong.

So the correct graph is the first one (the left - most graph).

Answer:

The left - most graph (the first graph among the three)