Problem 48 - 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. $$ s(x)=\frac{x+2}{(x+3)(x-1)} $$

Step-by-Step Solution

Answer

y-intercept: \((-\frac{2}{3}, 0)\), x-intercept: \((-2, 0)\); Vert. asymptotes: \(x = -3, x = 1\), Horiz. asymptote: \(y=0\). Domain: \(x \neq -3, 1\), Range: \(y \neq 0\).

Step 1: Find the y-intercept

The y-intercept of a function occurs when \(x = 0\). Substitute \(x = 0\) into the function: \[ s(0) = \frac{0 + 2}{(0 + 3)(0 - 1)} = \frac{2}{-3} = -\frac{2}{3}. \] Thus, the y-intercept is \((-\frac{2}{3}, 0)\).

Step 2: Find the x-intercepts

The x-intercepts occur where the numerator is zero and the denominator is not zero. Set the numerator equal to zero:\[ x + 2 = 0 \] Solve for \(x\):\[ x = -2. \] Thus, the x-intercept is \((-2, 0)\).

Step 3: Identify the vertical asymptotes

Vertical asymptotes occur where the denominator is zero (and the numerator is non-zero). Solve:\[ (x + 3)(x - 1) = 0 \] So, \(x + 3 = 0\) or \(x - 1 = 0\), resulting in solutions:\[ x = -3 \quad \text{and} \quad x = 1. \] The vertical asymptotes are at \(x = -3\) and \(x = 1\).

Step 4: Identify the horizontal asymptote

Compare the degrees of the numerator and the denominator. The degree of the numerator is 1, and the degree of the denominator is 2. Since the degree of the denominator is greater, the horizontal asymptote is:\[ y = 0. \]

Step 5: State the domain and range

The domain is all real numbers except where the denominator is zero: \(x eq -3\) and \(x eq 1\), so the domain is\( \{ x \mid x eq -3, 1 \} \). For the range, the function can take any value except the horizontal asymptote, \(y = 0\). Hence, the range is\( \{ y \mid y eq 0 \} \).

Step 6: Sketch the graph

Start by plotting the intercepts: the y-intercept \((-\frac{2}{3}, 0)\) and the x-intercept \((-2,0)\). Draw vertical lines at the asymptotes \(x = -3\) and \(x = 1\), indicating values the function cannot reach. The horizontal asymptote, \(y = 0\), shows how the graph behaves at extreme values of \(x\). The graph should approach the asymptotes but never touch them. Confirm the graph's accuracy using a graphing device.

Key Concepts

Intercepts

Intercepts are points where the graph of a function crosses the axes. For rational functions, these are important to find as they help in sketching the graph.

• Y-intercept: This occurs where the graph crosses the y-axis, which is at the point where the value of x is zero. For our function, substitute zero into the equation: \[ s(0) = \frac{0 + 2}{(0 + 3)(0 - 1)} = \frac{2}{-3} = -\frac{2}{3}. \] So, the y-intercept is at \((0, -\frac{2}{3})\).
• X-intercept: To find where the graph crosses the x-axis, we set the numerator equal to zero. This gives us: \[ x + 2 = 0, \] leading to \( x = -2 \). Thus, the x-intercept is \((-2, 0)\).

By identifying these intercepts, you have clear points where the graph intersects the axes. This is essential for accurate graph sketching.

Asymptotes

Asymptotes are lines that the graph approaches but never actually touches or crosses. There are two types for this rational function: vertical and horizontal.

• Vertical Asymptotes: These occur where the denominator of the function is zero. For \(s(x)\), solve \[ (x+3)(x-1) = 0, \]resulting in vertical asymptotes at \(x = -3\) and \(x = 1\). The graph will shoot up or down infinitely when it approaches these lines.
• Horizontal Asymptote: To find this, compare the degrees of the numerator and denominator. In this case, the degree of the numerator \((1)\) is less than that of the denominator \((2)\). Thus, there's a horizontal asymptote at \[ y = 0. \] This means as \(x\) becomes very large or very small, the graph will level out towards this line.

Understanding where asymptotes are helps to anticipate how the function behaves across its domain, providing a clear outline of the graph's general shape.

Domain and Range

The domain and range of a rational function describe the set of permissible x-values and y-values the function can take.

• Domain: This tells us all possible inputs \(x\) for the function. Here, the domain excludes values that make the denominator zero, which are \(x = -3\) and \(x = 1\). Hence, the domain is expressed as all real numbers except these: \[ \{ x \mid x eq -3, x eq 1 \}. \]
• Range: The range shows all possible outputs \(y\) of the function. For \(s(x)\), the range excludes the horizontal asymptote, \(y = 0\). This means every other real number is in the range, written as: \[ \{ y \mid y eq 0 \}. \]

Recognizing these features simplifies graphing as it provides boundaries and limitations, helping you graph the function correctly and understand which outputs are unattainable.