Problem 26 - Algebra and Trigonometry with Analytic Geometry

Question

Sketch the graph of \(f\). $$ f(x)=\frac{2 x^{2}+4 x-48}{x^{2}+3 x-10} $$

Step-by-Step Solution

Answer

The graph has vertical asymptotes at \(x = -5\) and \(x = 2\), a horizontal asymptote at \(y = 2\), x-intercepts at \(x = -6\) and \(x = 4\), and a y-intercept at \(y = 4.8\).

Step 1: Identify the Function's Domain

The domain of the function is the set of all real numbers except where the denominator is zero. Start by setting the denominator to 0 and solving for its roots: \[x^2 + 3x - 10 = 0\]. Factor: \[(x + 5)(x - 2) = 0\]. Thus, \(x = -5\) and \(x = 2\). Exclude \(x = -5\) and \(x = 2\) from the domain.

Step 2: Simplify the Function

Simplify \(f(x) = \frac{2x^2 + 4x - 48}{x^2 + 3x - 10}\) by factoring the numerator and denominator: - Numerator: \(2(x^2 + 2x - 24) = 2(x+6)(x-4)\)- Denominator: \((x+5)(x-2)\). Write the simplified function:\[f(x) = \frac{2(x+6)(x-4)}{(x+5)(x-2)}\].

Step 3: Determine Vertical Asymptotes

As vertical asymptotes occur where the denominator is zero (if these points aren't canceled by factors in the numerator), the vertical asymptotes are at \(x = -5\) and \(x = 2\).

Step 4: Locate the Holes

To locate the holes, check the canceled factors in the simplification step. Since no factor cancels out completely in the simplified version, there are no removable discontinuities or holes.

Step 5: Find Horizontal Asymptote

For horizontal asymptotes, analyze the degree of the polynomial. Both the numerator and the denominator are quadratic (same degree), thus the horizontal asymptote is \(y = \frac{2}{1} = 2\).

Step 6: Intersection Points with Axes

Set \(f(x) = 0\) to find the x-intercepts, solve:\[2(x+6)(x-4) = 0 \Rightarrow x = -6, 4\]. For y-intercept, set \(x = 0\), calculate:\[f(0) = \frac{2 \times 0^2 + 4 \times 0 - 48}{0^2 + 3 \times 0 - 10} = \frac{-48}{-10} = 4.8\]. Thus, y-intercept is \(y = 4.8\).

Step 7: Sketch the Graph

Use the analysis from the prior steps:- Draw vertical asymptotes at \(x = -5\) and \(x = 2\).- Plot the horizontal asymptote \(y = 2\).- Mark the x-intercepts at \(x = -6\) and \(x = 4\), and the y-intercept at \(y = 4.8\).- Sketch the curve, considering it approaches the asymptotes.

Key Concepts

Domain of a Function

Understanding the domain of a function is crucial because it tells us the set of all possible input values (x-values) for which the function is defined. For rational functions like the one given, the domain excludes values that make the denominator zero because division by zero is undefined.
To find these values, set the denominator equal to zero and solve for x:

• The function's denominator is \(x^2 + 3x - 10\).
• Factor it into \((x+5)(x-2)\).
• Setting each factor equal to zero gives: \(x + 5 = 0\) or \(x - 2 = 0\).
• The solutions are \(x = -5\) and \(x = 2\).

Thus, the domain of the function is all real numbers except for \(x = -5\) and \(x = 2\). This ensures we include only values that keep the function well-defined, avoiding undefined expressions in the calculations.

Asymptotes

Asymptotes are lines that the graph of the function approaches but never actually touches. They play a vital role in defining the end behavior of a function. For this exercise, we need to find both vertical and horizontal asymptotes.

Vertical Asymptotes

Vertical asymptotes appear where the denominator of a simplified rational function is zero, as long as these zeros are not canceled out by the numerator. Here:

• The simplified function is \(f(x) = \frac{2(x+6)(x-4)}{(x+5)(x-2)}\).
• The factors in the denominator are \((x+5)\) and \((x-2)\).
• These correspond to vertical asymptotes at \(x = -5\) and \(x = 2\).

Horizontal Asymptotes

These occur if the degrees of the numerator and denominator are the same, or if the degree of the numerator is less than the degree of the denominator. In this case:

• Both numerator and denominator are quadratic (degree of 2).
• The horizontal asymptote is found by dividing the leading coefficients: \(y = \frac{2}{1} = 2\).

Therefore, the horizontal asymptote is \(y = 2\). This guides how the function behaves far out along the x-axis.

X-intercepts and Y-intercepts

Intercepts are points where the graph crosses the axes, giving us valuable reference points for sketching the graph. Let's find both the x-intercepts and y-intercept for this rational function.

X-intercepts

X-intercepts are found by setting the numerator equal to zero and solving for x, as the output (y-value) is zero when crossing the x-axis. For our function:

• Use the numerator from the simplified form: \(2(x+6)(x-4)\).
• Set equal to zero: \(2(x+6)(x-4) = 0\).
• Solving gives \(x = -6\) and \(x = 4\).

Y-intercept

The y-intercept is located by setting \(x = 0\) and solving for \(f(x)\), since at the y-axis, the x-value is zero:

• Substitute \(x = 0\) into the simplified function.
• Calculate: \(f(0) = \frac{-48}{-10}\) which simplifies to \(y = 4.8\).

Thus, the y-intercept is at \(y = 4.8\). These intercepts provide exact locations where the graph intersects the axes, helping to build an accurate sketch of the graph.