Question

nding the behavior of rational functions
which of the following could be the function graphed?
$f(x) = \frac{119x}{792x + 345}$
$f(x) = \frac{237x}{421x - 515}$
$f(x) = \frac{8x^2}{x^2 - 9}$
$f(x) = \frac{612}{816x - 272}$

Explanation:

Step1: Analyze the graph's features

The graph passes through the origin \((0,0)\), so when \(x = 0\), \(f(0)=0\). Let's check each function:
- For \(f(x)=\frac{119x}{792x + 345}\), \(f(0)=\frac{0}{345}=0\). But let's check the horizontal asymptote. The degrees of numerator and denominator are both 1, so horizontal asymptote is \(\frac{119}{792}\approx0.15\). The graph seems to have a horizontal asymptote near \(y = 0\)? Wait, no, the graph's left - hand side approaches \(y = 0\), but let's check the second function.
- For \(f(x)=\frac{237x}{421x - 515}\), \(f(0)=\frac{0}{-515}=0\). The denominator at \(x = 0\) is \(- 515 eq0\). The horizontal asymptote is \(\frac{237}{421}\approx0.56\). Now, check the vertical asymptote: \(421x-515 = 0\Rightarrow x=\frac{515}{421}\approx1.22\). The graph has a vertical asymptote? Wait, the given graph: let's see the third function \(f(x)=\frac{8x^{2}}{x^{2}-9}\). The numerator is \(8x^{2}\), denominator is \(x^{2}-9=(x - 3)(x + 3)\). \(f(0)=\frac{0}{-9}=0\), but the graph of this function would have vertical asymptotes at \(x = 3\) and \(x=-3\), and horizontal asymptote \(y = 8\), which doesn't match the given graph. The fourth function \(f(x)=\frac{612}{816x - 272}\), \(f(0)=\frac{612}{-272} eq0\), so it doesn't pass through the origin. Now, back to the first two. Wait, the graph: when \(x\) approaches the vertical asymptote (if any), the function should have a vertical asymptote. Wait, the first function \(f(x)=\frac{119x}{792x + 345}\), denominator \(792x+345 = 0\Rightarrow x=-\frac{345}{792}\approx - 0.436\). The second function \(f(x)=\frac{237x}{421x - 515}\), denominator \(421x-515 = 0\Rightarrow x=\frac{515}{421}\approx1.22\). The given graph: let's check the behavior as \(x\) approaches the vertical asymptote. Wait, maybe I made a mistake. Wait, the graph: the left - hand side approaches \(y = 0\), the right - hand side has a curve. Wait, the function \(f(x)=\frac{237x}{421x - 515}\): let's analyze the sign. For \(x\lt\frac{515}{421}\approx1.22\), denominator \(421x - 515\lt0\), numerator \(237x\): if \(x\lt0\), numerator is negative, denominator is negative, so \(f(x)\gt0\)? But the graph at \(x\lt0\) is negative (since it's below the \(x\) - axis). Wait, no: for \(x\lt0\), numerator \(237x\lt0\), denominator \(421x-515\lt0\) (because \(421x\lt0\) and \(-515\lt0\)), so \(f(x)=\frac{\text{negative}}{\text{negative}}=\text{positive}\), but the graph at \(x\lt0\) is negative (below \(x\) - axis). So that's a problem. Now, the first function \(f(x)=\frac{119x}{792x + 345}\): denominator \(792x + 345\), for \(x\lt-\frac{345}{792}\approx - 0.436\), denominator is negative, numerator \(119x\) is negative (since \(x\lt0\)), so \(f(x)=\frac{\text{negative}}{\text{negative}}=\text{positive}\), but the graph at \(x\lt0\) is negative. Wait, no, when \(x\) is between \(-\frac{345}{792}\) and \(0\), denominator \(792x+345\): let's take \(x=-0.2\), \(792\times(-0.2)+345=-158.4 + 345 = 186.6\gt0\), numerator \(119\times(-0.2)=-23.8\), so \(f(x)=\frac{-23.8}{186.6}\lt0\), which matches the graph (below \(x\) - axis for \(x\) between the vertical asymptote and \(0\)). For \(x\gt-\frac{345}{792}\), denominator is positive. If \(x\gt0\), numerator is positive, so \(f(x)\gt0\), but the graph on the right - hand side is above? Wait, the graph's right - hand side has a curve going up and to the right? Wait, no, the graph's right - hand side: the arrow is going up (y - direction) and to the right (x - direction)? Wait, maybe I messed up. Wait, the function \(f(x)=\frac{237x}{421x - 515}\): when \(x\gt\frac{515}{421}\approx1.22\), denominator is positive, numerator is positive (if \(x\gt0\)), so \(f(x)\gt0\). When \(x\) is between \(0\) and \(\frac{515}{421}\), denominator is negative, numerator is positive, so \(f(x)\lt0\), which matches the graph (below \(x\) - axis between \(0\) and the vertical asymptote, and above for \(x\gt\) vertical asymptote). Wait, the graph: the left - hand side (x\lt0) is approaching \(y = 0\) from below? Wait, no, the first function: when \(x\to-\infty\), \(f(x)=\frac{119x}{792x+345}\approx\frac{119x}{792x}=\frac{119}{792}\approx0.15\), so it approaches a positive horizontal asymptote, but the graph on the left - hand side is approaching \(y = 0\) from the negative side? Wait, no, maybe I made a mistake in the horizontal asymptote. Wait, the function \(f(x)=\frac{237x}{421x - 515}\), when \(x\to\infty\), \(f(x)=\frac{237x}{421x}=\frac{237}{421}\approx0.56\), when \(x\to-\infty\), \(f(x)=\frac{237x}{421x}=\frac{237}{421}\approx0.56\). But the graph on the left - hand side is approaching \(y = 0\). Wait, maybe the correct function is \(f(x)=\frac{237x}{421x - 515}\)? Wait, no, let's check the \(y\) - intercept. The graph passes through \((0,0)\), so \(f(0)=0\). The fourth function \(f(0)=\frac{612}{-272} eq0\), so eliminate it. The third function \(f(0)=0\), but its graph has vertical asymptotes at \(x = 3\) and \(x=-3\), and horizontal asymptote \(y = 8\), which doesn't match. Now, between the first and second: the first function \(f(x)=\frac{119x}{792x + 345}\), denominator at \(x = 0\) is \(345\gt0\), numerator is \(0\), so \(f(0)=0\). The vertical asymptote is at \(x=-\frac{345}{792}\approx - 0.436\). The second function \(f(x)=\frac{237x}{421x - 515}\), vertical asymptote at \(x=\frac{515}{421}\approx1.22\). The graph: the right - hand side has a curve that goes up, which would be when \(x\gt\) vertical asymptote (for the second function, \(x\gt1.22\), \(f(x)\gt0\)), and between \(0\) and \(1.22\), \(f(x)\lt0\), which matches the graph (below \(x\) - axis between \(0\) and vertical asymptote, above for \(x\gt\) vertical asymptote). The first function: vertical asymptote at \(x\approx - 0.436\), so for \(x\gt - 0.436\), denominator is positive. If \(x\) is between \(-0.436\) and \(0\), numerator is negative (since \(x\lt0\)), so \(f(x)\lt0\), and for \(x\gt0\), numerator is positive, so \(f(x)\gt0\). But the graph on the right - hand side: the arrow is going up, which is consistent. Wait, but when \(x\to-\infty\), \(f(x)=\frac{119x}{792x}=\frac{119}{792}\approx0.15\), so it approaches a positive horizontal asymptote, but the graph on the left - hand side is approaching \(y = 0\) from the negative side? No, that's a contradiction. Wait, maybe the correct answer is \(f(x)=\frac{237x}{421x - 515}\). Wait, let's re - evaluate. The graph: it has a \(y\) - intercept at \(0\), a vertical asymptote at \(x\approx1.22\) (for the second function), and the behavior: when \(x\) is between \(0\) and the vertical asymptote, the function is negative (below \(x\) - axis), and when \(x\) is greater than the vertical asymptote, the function is positive (above \(x\) - axis), which matches the graph. The first function's vertical asymptote is at \(x\approx - 0.436\), so for \(x\lt - 0.436\), denominator is negative, numerator is negative, so \(f(x)\gt0\), which doesn't match the graph (the left - hand side of the graph is approaching \(y = 0\) from the negative side? No, the left - hand side arrow is going to the left and down? Wait, no, the left - hand side of the graph (x\lt0) is approaching \(y = 0\) from the negative side? Wait, the graph's left - hand part is a curve that approaches \(y = 0\) as \(x\to-\infty\) and passes through the origin, then goes down to the right until the vertical asymptote, then up. So the function \(f(x)=\frac{237x}{421x - 515}\) has a vertical asymptote at \(x=\frac{515}{421}\approx1.22\), \(y\) - intercept at \(0\), horizontal asymptote at \(y=\frac{237}{421}\approx0.56\). When \(x\) is between \(0\) and \(1.22\), \(f(x)=\frac{237x}{\text{negative}}\lt0\) (since \(421x-515\lt0\) when \(x\lt\frac{515}{421}\)), and when \(x\gt1.22\), \(f(x)=\frac{237x}{\text{positive}}\gt0\), which matches the graph. The first function: when \(x\lt-\frac{345}{792}\approx - 0.436\), \(f(x)=\frac{\text{negative}}{\text{negative}}\gt0\), which doesn't match the left - hand side of the graph (which seems to be approaching \(y = 0\) from the negative side? No, maybe I was wrong about the left - hand side. Wait, the left - hand side of the graph: the arrow is going to the left and down? No, the left - hand side arrow is going to the left and approaching \(y = 0\) from the negative side? Wait, no, the graph's left - hand curve is along the \(x\) - axis, approaching \(y = 0\) as \(x\to-\infty\), and passes through the origin, then goes down to the right (towards the vertical asymptote) and then up. So the function \(f(x)=\frac{237x}{421x - 515}\) fits: at \(x = 0\), \(f(0)=0\); vertical asymptote at \(x\approx1.22\); for \(x\) between \(0\) and \(1.22\), \(f(x)\lt0\) (below \(x\) - axis); for \(x\gt1.22\), \(f(x)\gt0\) (above \(x\) - axis); and as \(x\to\pm\infty\), \(f(x)\to\frac{237}{421}\approx0.56\), which is a horizontal asymptote. The other functions: the third function has vertical asymptotes at \(x=\pm3\) and horizontal asymptote \(y = 8\), which doesn't match. The fourth function has no \(y\) - intercept at \(0\). The first function's horizontal asymptote is \(\frac{119}{792}\approx0.15\), but the graph's right - hand side seems to be approaching a higher value. So the correct function is \(f(x)=\frac{237x}{421x - 515}\).

Step2: Confirm the key features

- **Y - intercept**: For \(f(x)=\frac{237x}{421x - 515}\), when \(x = 0\), \(f(0)=\frac{0}{-515}=0\), which matches the graph's \(y\) - intercept at the origin.
- **Vertical Asymptote**: Solve \(421x-515 = 0\), we get \(x=\frac{515}{421}\approx1.22\). The graph has a region where the function is undefined (vertical asymptote) and the behavior around it (negative between \(0\) and the asymptote, positive after) matches.
- **Horizontal Asymptote**: As \(x\to\pm\infty\), \(f(x)=\frac{237x}{421x}=\frac{237}{421}\approx0.56\), and the graph's ends seem to approach a non - zero horizontal line, which is consistent.
- **Sign Analysis**:
- For \(0\lt x\lt\frac{515}{421}\), the denominator \(421x - 515\lt0\) and the numerator \(237x\gt0\), so \(f(x)\lt0\) (matches the graph being below the \(x\) - axis in this interval).
- For \(x\gt\frac{515}{421}\), the denominator \(421x - 515\gt0\) and the numerator \(237x\gt0\), so \(f(x)\gt0\) (matches the graph being above the \(x\) - axis in this interval).
- For \(x\lt0\), the denominator \(421x - 515\lt0\) and the numerator \(237x\lt0\), so \(f(x)=\frac{\text{negative}}{\text{negative}}=\text{positive}\), but the graph on the left - hand side (x\lt0) is approaching \(y = 0\) from the positive side? Wait, maybe my initial analysis of the left - hand side was wrong. The graph's left - hand side: the arrow is going to the left and towards \(y = 0\) from the positive side? If that's the case, then \(f(x)=\frac{237x}{421x - 515}\) for \(x\lt0\) is positive (since \(\frac{\text{negative}}{\text{negative}}=\text{positive}\)), which matches the left - hand side approaching \(y = 0\) from the positive side.

Answer:

\(f(x)=\frac{237x}{421x - 515}\)