Problem 57 - Algebra and Trigonometry

Question

Find the intercepts and asymptotes, and then sketch a graph of the rational function and state the domain and range. Use a graphing device to confirm your answer. $$ r(x)=\frac{2 x^{2}+10 x-12}{x^{2}+x-6} $$

Step-by-Step Solution

Answer

The x-intercepts are \(-6\) and \(1\); y-intercept is 2. Vertical asymptotes are at \(x = -3\) and \(x = 2\). The horizontal asymptote is \(y = 2\).

Step 1: Factoring the numerator and denominator

First, factor the numerator \(2x^2 + 10x - 12\) and the denominator \(x^2 + x - 6\). For the numerator: \[ 2x^2 + 10x - 12 = 2(x^2 + 5x - 6) \] Factor further: \[ x^2 + 5x - 6 = (x + 6)(x - 1) \] So, the expression becomes \[ 2(x + 6)(x - 1) \] For the denominator:\[ x^2 + x - 6 = (x + 3)(x - 2) \]The function thus simplifies to: \[ \frac{2(x + 6)(x - 1)}{(x + 3)(x - 2)} \]

Step 2: Finding Intercepts

To find the x-intercepts, set the numerator equal to zero:\[ 2(x + 6)(x - 1) = 0 \]Solving this, we get \(x = -6\) and \(x = 1\). These are the x-intercepts of the graph.To find the y-intercept, evaluate \(r(0)\):\[ r(0) = \frac{2(0 + 6)(0 - 1)}{(0 + 3)(0 - 2)} = \frac{2 \cdot 6 \cdot (-1)}{3 \cdot (-2)} = 2 \]So, the y-intercept is \((0, 2)\).

Step 3: Finding Vertical Asymptotes

Vertical asymptotes occur where the denominator equals zero (but not the numerator):\[ (x + 3)(x - 2) = 0 \]Solving this gives \(x = -3\) and \(x = 2\) as the vertical asymptotes. Since these are not canceled by the numerator, they are true asymptotes.

Step 4: Finding Horizontal Asymptotes

For horizontal asymptotes, compare the degrees of the polynomials.Both the numerator and the denominator are degree 2, so the horizontal asymptote is:\[ y = \frac{2}{1} = 2 \]

Step 5: Analyzing Domain and Range

The domain consists of all real numbers except where the denominator is zero: \(x eq -3\), \(x eq 2\).Since the graph approaches the horizontal asymptote but never reaches it, the range is all real numbers except \(y eq 2\).

Step 6: Sketching the Graph

Plot the vertical asymptotes at \(x = -3\) and \(x = 2\) as dashed lines.Sketch the horizontal asymptote at \(y = 2\). Plot the intercepts \((-6, 0)\), \((1, 0)\), and \((0, 2)\) on the graph. The function will approach but not touch the asymptotes.Use a graphing device to confirm that the plot matches this description.

Key Concepts

Intercepts

Intercepts in a rational function are where the graph crosses the axes. To find the x-intercepts, you need to set the numerator equal to zero and solve for \(x\). In this exercise, the x-intercepts are at \(x = -6\) and \(x = 1\); this is because solving \(2(x + 6)(x - 1) = 0\) yields those solutions. These points are where the graph of the function will intersect the x-axis, represented as \((-6, 0)\) and \((1, 0)\).
For the y-intercept, substitute zero for \(x\) in the function \(r(x) = \frac{2(0 + 6)(0 - 1)}{(0 + 3)(0 - 2)}\). This calculation gives \(r(0) = 2\), so the y-intercept is at \((0, 2)\), which is where the graph crosses the y-axis. Understanding intercepts is crucial for sketching the graph as they are fixed points where the graph interacts with the coordinate plane axes.

Asymptotes

Asymptotes are lines that the graph of the function approaches but never actually touches. These can be of three types: vertical, horizontal, and oblique. For this rational function, we identify vertical and horizontal asymptotes.

• Vertical Asymptotes: Occur where the denominator is zero, indicating where the function is undefined. For the given function, solve \((x + 3)(x - 2) = 0\) to find \(x = -3\) and \(x = 2\). These x-values are vertical asymptotes since they aren't canceled out by terms in the numerator.
  
• Horizontal Asymptotes: Determined by comparing the degrees of the numerator and denominator. Both polynomials here are degree 2, so the horizontal asymptote is found by dividing the leading coefficients, resulting in \(y = 2\).
  

These asymptotes indicate how the graph behaves at extreme values of \(x\) and ensure it never crosses these lines, guiding the ultimate shape of the graph.

Domain and Range

The domain of a rational function consists of all values of \(x\) that keep the function defined, meaning the values for which the denominator does not equal zero. In this function, the denominator factors as \((x + 3)(x - 2)\), so the domain excludes \(x = -3\) and \(x = 2\), places where the function would be undefined.
The range concerns the possible values \(r(x)\) can take. Here, because of the horizontal asymptote, the function cannot equal \(y = 2\), making the range all real numbers except \(y = 2\). It's essential to note the reasons for exclusions in the domain and the impact of asymptotes in limiting the range. Both concepts are key to understanding how the graph portrays all potential function outputs and inputs.

Graphing

When graphing a rational function such as this one, following a methodical approach helps create an accurate sketch. Start by

• plotting the intercepts: \((-6, 0)\), \((1, 0)\) (x-intercepts), and \((0, 2)\) (y-intercept).
  
• Identify and sketch the asymptotes as dashed lines: vertical asymptotes as \(x = -3\) and \(x = 2\) and a horizontal asymptote as \(y = 2\). These lines guide the overall direction of the curves in the graph.
  
• Sketch the typical behavior of the curve approaching but not meeting the asymptotes. Check that the graph aligns with the intercepts plotted earlier and follows the general shape indicated by any asymptotes.
  

Using a graphing calculator or software can help confirm the accuracy of your sketch, allowing you to visualize how the rational function behaves across the domain visually.