We test the sign of $ rac{-3x + 11}{x - 2} $ in each interval:
["# Testing the Sign of $ \dfrac{-3x + 11}{x - 2} $ in Each Interval", "When analyzing rational functions, one of the key tasks is determining the sign (positive or negative) of the function in different intervals defined by its zeros and points of undefinedness. In this article, we explore how to test the sign of the rational expression\n$$\nf(x) = \dfrac{-3x + 11}{x - 2}\n$$\nacross each interval created by its critical points. Understanding the sign helps shape the behavior of the function, important for graphing, solving inequalities, and solving real-world problems.", "---", "## Step 1: Identify the Zeros and Vertical Asymptote", "First, find where the numerator and denominator are zero:", "- Numerator zero:\n $$\n -3x + 11 = 0 \Rightarrow x = \dfrac{11}{3} \approx 3.67\n $$", "- Denominator zero:\n $$\n x - 2 = 0 \Rightarrow x = 2\n $$", "The function is undefined at $ x = 2 $ — this is a vertical asymptote. Also, the numerator is zero at $ x = \dfrac{11}{3} $, where the function crosses or touches the x-axis (since the denominator is not zero there).", "These critical points divide the real number line into three intervals:", "1. $ (-\infty, 2) $\n2. $ \left(2,\ \dfrac{11}{3}\right) $\n3. $ \left(\dfrac{11}{3},\ \infty\right) $", "We will test the sign of $ f(x) $ in each interval.", "---", "## Step 2: Choose Test Points", "Pick a simple number from each interval and substitute into $ f(x) $:", "- Interval $ (-\infty, 2) $: Try $ x = 0 $\n- Interval $ \left(2,\ \dfrac{11}{3}\right) $: Try $ x = 3 $\n- Interval $ \left(\dfrac{11}{3},\ \infty\right) $: Try $ x = 4 $", "---", "## Step 3: Evaluate the Sign of Each Factor", "Evaluate numerator $ -3x + 11 $ and denominator $ x - 2 $:", "- At $ x = 0 $:\n - Numerator: $ -3(0) + 11 = 11 > 0 $\n - Denominator: $ 0 - 2 = -2 < 0 $\n - So, $ f(0) = \dfrac{+}{-} = - $", "- At $ x = 3 $:\n - Numerator: $ -3(3) + 11 = -9 + 11 = 2 > 0 $\n - Denominator: $ 3 - 2 = 1 > 0 $\n - So, $ f(3) = \dfrac{+}{+} = + $", "- At $ x = 4 $:\n - Numerator: $ -3(4) + 11 = -12 + 11 = -1 < 0 $\n - Denominator: $ 4 - 2 = 2 > 0 $\n - So, $ f(4) = \dfrac{-}{+} = - $", "---", "## Step 4: Conclusion on the Sign of $ f(x) $", "| Interval | Testing Point | $ -3x + 11 $ | $ x - 2 $ | Sign of $ f(x) $ |\n|------------------|---------------|----------------|------------|--------------------|\n| $ (-\infty, 2) $ | $ x = 0 $ | + | – | – (negative) |\n| $ \left(2, \dfrac{11}{3}\right) $ | $ x = 3 $ | + | + | + (positive) |\n| $ \left(\dfrac{11}{3}, \infty\right) $ | $ x = 4 $ | – | + | – (negative) |", "---", "## Step 5: Behavior at Critical Points", "- At $ x = \dfrac{11}{3} $, $ f(x) = 0 $, since the numerator is zero and denominator is non-zero.\n- At $ x = 2 $, the function is undefined (vertical asymptote), so exclude this point.", "---", "## Why This Matters", "Knowing the sign of $ \dfrac{-3x + 11}{x - 2} $ helps:", "- Identify where the function is positive or negative — useful in equations $ f(x) > 0 $ or $ f(x) < 0 $ solving.\n- Sketch the graph accurately, showing where the function rises (positive) or falls (negative).\n- Application in optimization, economics, physics, and modeling scenarios where ratios define behavior.", "---", "## Summary", "- The function $ f(x) = \dfrac{-3x + 11}{x - 2} $ has a zero at $ x = \dfrac{11}{3} $ and a vertical asymptote at $ x = 2 $.\n- In $ (-\infty, 2) $, $ f(x) < 0 $\n- In $ \left(2,\ \dfrac{11}{3}\right) $, $ f(x) > 0 $\n- In $ \left(\dfrac{11}{3},\ \infty\right) $, $ f(x) < 0 $", "This sign analysis is foundational when working with rational functions and understanding their complete behavior across the number line.", "---", "### Further Reading\n- How to analyze rational functions\n- Sign charts for inequalities\n- Graphing rational functions step-by-step", "---", "Keywords: rational function, sign of a function, test intervals, $ \dfrac{-3x + 11}{x - 2} $, sign chart, critical points, vertical asymptote, function behavior, algebra tutorial"]
