Match the following f(x)=(-3x-1)/(3x+2) f(x)=(-8x+3)/(4x) f(x)=(x+4)/(x-5) f(x)=(6)/(x-1) f(x)=(6x+1)/(3x-5) HA at x-axis HA at x=1 HA at x=-1 HA at x=2 HA at x=-2

To match the given functions with their horizontal asymptotes (HA), we need to analyze the degrees of the numerator and denominator for each function. The horizontal asymptote depends on the highest power of \( x \) in both the numerator and the denominator.<br /><br />1. \( f(x) = \frac{-3x-1}{3x+2} \)<br /> - Both the numerator and the denominator are linear (degree 1).<br /> - The horizontal asymptote is the ratio of the leading coefficients: \( \frac{-3}{3} = -1 \).<br /> - **HA at x-axis** (y = -1)<br /><br />2. \( f(x) = \frac{-8x+3}{4x} \)<br /> - The numerator is linear (degree 1) and the denominator is linear (degree 1).<br /> - The horizontal asymptote is the ratio of the leading coefficients: \( \frac{-8}{4} = -2 \).<br /> - **HA at x = -2**<br /><br />3. \( f(x) = \frac{x+4}{x-5} \)<br /> - Both the numerator and the denominator are linear (degree 1).<br /> - The horizontal asymptote is the ratio of the leading coefficients: \( \frac{1}{1} = 1 \).<br /> - **HA at x-axis** (y = 1)<br /><br />4. \( f(x) = \frac{6}{x-1} \)<br /> - The numerator is constant (degree 0) and the denominator is linear (degree 1).<br /> - The horizontal asymptote is \( y = 0 \) (since the degree of the denominator is greater than the degree of the numerator).<br /> - **HA at x-axis** (y = 0)<br /><br />5. \( f(x) = \frac{6x+1}{3x-5} \)<br /> - Both the numerator and the denominator are linear (degree 1).<br /> - The horizontal asymptote is the ratio of the leading coefficients: \( \frac{6}{3} = 2 \).<br /> - **HA at x = 2**<br /><br />So, the matching is as follows:<br /><br />1. \( f(x) = \frac{-3x-1}{3x+2} \) - HA at x-axis<br />2. \( f(x) = \frac{-8x+3}{4x} \) - HA at \( x = -2 \)<br />3. \( f(x) = \frac{x+4}{x-5} \) - HA at x-axis<br />4. \( f(x) = \frac{6}{x-1} \) - HA at x-axis<br />5. \( f(x) = \frac{6x+1}{3x-5} \) - HA at \( x = 2 \)