Question

consider the function $f(x) = \frac{x^2 - 3x}{x^2 - 9}$. which graph is the graph of function $f$? \
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a. \
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graph a \
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b. \
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graph b

Explanation:

Step1: Simplify the function

First, factor the numerator and the denominator. The numerator \(x^2 - 3x = x(x - 3)\), and the denominator \(x^2 - 9=(x - 3)(x + 3)\) (using the difference of squares formula \(a^2 - b^2=(a - b)(a + b)\)). So we can simplify the function \(f(x)=\frac{x(x - 3)}{(x - 3)(x + 3)}\) for \(x eq3\) (since we can't divide by zero, and also \(x eq - 3\) because the denominator is zero when \(x=-3\)). Canceling out the common factor \((x - 3)\) (for \(x eq3\)), we get \(f(x)=\frac{x}{x + 3}\), with domain \(x eq - 3\) and \(x eq3\).

Step2: Analyze the vertical asymptote

The vertical asymptote occurs where the denominator is zero (after canceling common factors). So for \(f(x)=\frac{x}{x + 3}\), the denominator is zero when \(x=-3\). So the vertical asymptote is \(x = - 3\).

Step3: Analyze the horizontal asymptote

For a rational function \(f(x)=\frac{ax^n+...}{bx^m+...}\), if \(n = m\), the horizontal asymptote is \(y=\frac{a}{b}\). Here, the degree of the numerator (degree 1) and the degree of the denominator (degree 1) are equal. So the horizontal asymptote is \(y=\frac{1}{1}=1\).

Step4: Analyze the hole

A hole occurs where the canceled factor is zero, i.e., when \(x - 3=0\) or \(x = 3\). To find the \(y\)-coordinate of the hole, substitute \(x = 3\) into the simplified function \(f(x)=\frac{x}{x + 3}\). So \(f(3)=\frac{3}{3 + 3}=\frac{3}{6}=\frac{1}{2}\). Wait, no, wait. Wait, when we simplified, we had \(f(x)=\frac{x}{x + 3}\) for \(x eq3\) and \(x eq - 3\). Wait, actually, the original function is undefined at \(x = 3\) and \(x=-3\). The hole is at \(x = 3\) (since that's where the common factor was canceled), and the vertical asymptote is at \(x=-3\).

Now let's analyze the graphs:

- For graph A: The vertical asymptote seems to be at \(x = 3\) (the dashed line is at \(x = 3\)), which is wrong because our vertical asymptote is at \(x=-3\). Wait, no, wait, maybe I misread. Wait, let's check the graphs again.

Wait, graph B has a vertical asymptote at \(x=-3\) (the dashed line is at \(x=-3\)), and graph A has a vertical asymptote at \(x = 3\). Let's check the horizontal asymptote. The horizontal asymptote is \(y = 1\). Let's check the behavior as \(x\to\pm\infty\). For \(f(x)=\frac{x}{x + 3}=\frac{1}{1+\frac{3}{x}}\), as \(x\to\infty\), \(\frac{3}{x}\to0\), so \(f(x)\to1\), and as \(x\to-\infty\), \(\frac{3}{x}\to0\), so \(f(x)\to1\). So the horizontal asymptote is \(y = 1\).

Now, let's check the hole. The hole is at \(x = 3\), so we should have an open circle at \(x = 3\). Let's check the \(y\)-value at \(x = 3\) in the simplified function: \(f(3)=\frac{3}{3 + 3}=\frac{1}{2}\)? Wait, no, wait, no: Wait, the simplified function is \(f(x)=\frac{x}{x + 3}\), so at \(x = 3\), \(f(3)=\frac{3}{6}=\frac{1}{2}\)? Wait, but earlier when we factored, the original function was \(\frac{x(x - 3)}{(x - 3)(x + 3)}\), so at \(x = 3\), both numerator and denominator are zero, so it's a hole. So the hole is at \((3,\frac{3}{3 + 3})=(3,\frac{1}{2})\)? Wait, no, wait, no, I think I made a mistake. Wait, the simplified function is \(f(x)=\frac{x}{x + 3}\) for \(x eq3\) and \(x eq - 3\). So when \(x = 3\), the original function is undefined, but the simplified function gives \(f(3)=\frac{3}{6}=\frac{1}{2}\), so the hole is at \((3,\frac{1}{2})\). But let's check the graphs.

Wait, graph B: Let's see the vertical asymptote is at \(x=-3\) (the dashed line is at \(x=-3\)), horizontal asymptote \(y = 1\) (the dashed horizontal line). The hole: when \(x = 3\), let's see the graph. Wait, maybe I messed up the vertical asymptote. Wait, no, the denominator of the simplified function is \(x + 3\), so vertical asymptote at \(x=-3\). So graph B has vertical asymptote at \(x=-3\), horizontal asymptote \(y = 1\), and a hole? Wait, no, the original function has a hole at \(x = 3\) and vertical asymptote at \(x=-3\). Let's check the behavior of the function.

For \(x>-3\) and \(x eq3\): Let's take \(x = 0\), \(f(0)=\frac{0}{0 + 3}=0\). So the function passes through \((0,0)\) when \(x eq3\) and \(x eq - 3\). Let's check graph B: when \(x = 0\), the graph of B: the lower part (for \(x>0\)) seems to approach the horizontal asymptote \(y = 1\) from below, and the upper part (for \(x<-3\)): let's take \(x=-4\), \(f(-4)=\frac{-4}{-4 + 3}=\frac{-4}{-1}=4\). So at \(x=-4\), \(f(-4)=4\). Let's check graph B: the upper part (left of \(x=-3\)) has a point near \(x=-4\) with \(y = 4\), which matches. Graph A: vertical asymptote at \(x = 3\), which is wrong. So the correct graph should be B? Wait, no, wait, the original function is \(f(x)=\frac{x^2-3x}{x^2 - 9}=\frac{x(x - 3)}{(x - 3)(x + 3)}\), so after canceling, \(f(x)=\frac{x}{x + 3}\), \(x eq3,-3\). So the vertical asymptote is \(x=-3\) (denominator of simplified function), hole at \(x = 3\) (where the canceled factor is zero). The horizontal asymptote is \(y = 1\) (since degrees of numerator and denominator are equal, ratio of leading coefficients is 1/1). Now, let's check the sign of the function. For \(x>-3\) and \(x eq3\): if \(x\in(-3,3)\), say \(x = 0\), \(f(0)=0\); if \(x>3\), say \(x = 4\), \(f(4)=\frac{4}{4 + 3}=\frac{4}{7}\approx0.57\), which is positive and less than 1. For \(x<-3\), say \(x=-4\), \(f(-4)=\frac{-4}{-4 + 3}=\frac{-4}{-1}=4\), which is positive and greater than 1. Now, looking at graph B: left of \(x=-3\) (vertical asymptote at \(x=-3\)), the graph goes up (positive and increasing towards the asymptote), right of \(x=-3\) and left of \(x = 3\) (hole at \(x = 3\)), the graph passes through (0,0) (since \(f(0)=0\)), and right of \(x = 3\), the graph approaches \(y = 1\) from below (since \(f(4)=\frac{4}{7}<1\)). Graph A has vertical asymptote at \(x = 3\), which is incorrect because the vertical asymptote should be at \(x=-3\). So the correct graph is B.

Answer:

B. The graph of function \(f\) (the graph labeled B)