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Graph the rational function. \[ f(x)=\frac{-x+5}{-2 x+1} \] Start by drawing the vertical and horizontal asymptotes. T

Ask by Gordon Alexander. in the United States
Mar 14,2025

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**Vertical Asymptote:** \( x = \frac{1}{2} \) **Horizontal Asymptote:** \( y = \frac{1}{2} \)

Solución

**Step 1. Find the Vertical Asymptote** The vertical asymptote is determined by setting the denominator equal to zero. \[ -2x + 1 = 0 \] Solve for \( x \): \[ -2x = -1 \quad \Rightarrow \quad x = \frac{1}{2} \] So, the vertical asymptote is: \[ x = \frac{1}{2} \] **Step 2. Find the Horizontal Asymptote** For rational functions where the degrees of the numerator and denominator are equal, the horizontal asymptote is determined by the ratio of the leading coefficients. The numerator is \(-x + 5\) (leading coefficient \(-1\)) and the denominator is \(-2x + 1\) (leading coefficient \(-2\)). Thus, the horizontal asymptote is: \[ y = \frac{-1}{-2} = \frac{1}{2} \] So, the horizontal asymptote is: \[ y = \frac{1}{2} \] **Summary** - **Vertical Asymptote:** \( x = \frac{1}{2} \) - **Horizontal Asymptote:** \( y = \frac{1}{2} \) These lines should be drawn on the graph to begin sketching the function \( f(x) = \frac{-x+5}{-2x+1} \).

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To graph the rational function \( f(x) = \frac{-x+5}{-2x+1} \), begin by finding the vertical asymptote. Set the denominator \(-2x + 1 = 0\), which gives \(x = \frac{1}{2}\). This is where the function will be undefined and thus you can draw a dashed vertical line at \(x = \frac{1}{2}\). Next, find the horizontal asymptote by looking at the leading coefficients of the numerator and the denominator. Since both have a degree of 1, the horizontal asymptote is determined by the ratio of the leading coefficients, which is \(\frac{-1}{-2} = \frac{1}{2}\). Draw a dashed horizontal line at \(y = \frac{1}{2}\). Now you have your asymptotes! Let's make the function come to life with those lines in place. Plot the intercepts by setting \(x=0\) to find the y-intercept: \(f(0) = \frac{5}{1} = 5\). That's a solid point at \((0, 5)\). For the x-intercept, set the numerator to zero: \(-x + 5 = 0\) gives \(x=5\), resulting in the point \((5, 0)\). Now sketch the curve by approaching the asymptotes and connecting the intercepts smoothly. You'll see it heads towards the vertical asymptote at \(x=\frac{1}{2}\) and flattens out near the horizontal asymptote as \(x\) approaches infinity. Happy graphing!

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