Problem 37 - Precalculus: Functions and Graphs

Question

Simplify \(f(x),\) and sketch the graph of \(f\) $$f(x)=\frac{2 x^{2}+x-6}{x^{2}+3 x+2}$$

Step-by-Step Solution

Answer

The simplified function is \(f(x) = \frac{2x - 3}{x + 1}\) with a hole at \(x = -2\) and a vertical asymptote at \(x = -1\).

Step 1: Factor the Numerator

Let's start by factoring the numerator of the given function. The numerator is a quadratic expression: \(2x^2 + x - 6\). We need to find two numbers that multiply to give us \(-12\) (which is \(2 \cdot -6\)) and add to give us \(1\) (the coefficient of \(x\)). These numbers are \(4\) and \(-3\). Thus, we factor the numerator as \((2x - 3)(x + 2)\).

Step 2: Factor the Denominator

Next, we will factor the denominator \(x^2 + 3x + 2\). We need two numbers that multiply to \(2\) and add to \(3\). These numbers are \(1\) and \(2\). Thus, the denominator factors into \((x + 1)(x + 2)\).

Step 3: Simplify the Function

Now that we have both the numerator and the denominator factored, \(f(x)\) becomes: \[ f(x) = \frac{(2x - 3)(x + 2)}{(x + 1)(x + 2)} \]. Notice that the \((x + 2)\) terms in the numerator and denominator cancel out, simplifying the function to: \[ f(x) = \frac{2x - 3}{x + 1} \].

Step 4: Determine the Domain

The domain of the simplified function is restricted by the values that make the original denominator zero. The original denominator \(x^2 + 3x + 2\) is zero for \(x = -1\) and \(x = -2\). Therefore, the domain of \(f(x)\) is all real numbers except \(x = -1\) and \(x = -2\).

Step 5: Identify Discontinuities

Although \((x + 2)\) cancels out, this means there is a removable discontinuity at \(x = -2\). There is an asymptote at \(x = -1\) because \((x + 1)\) remains in the denominator of the simplified function.

Step 6: Sketch the Graph

To sketch the graph of the simplified function \(f(x) = \frac{2x - 3}{x + 1}\), note that there is a vertical asymptote at \(x = -1\) and a removable discontinuity (hole) at \(x = -2\). The x-intercept occurs when the numerator is zero, \(2x - 3 = 0\), giving \(x = \frac{3}{2}\). The horizontal asymptote is given by comparing the degrees of the numerator and denominator, since both are degree 1, it occurs at \(y = 2\) (leading coefficient ratio). Plot these features to sketch the graph accurately.

Key Concepts

Factoring Polynomials

Factoring polynomials is an essential step in simplifying rational functions. It involves expressing a polynomial as the product of its factors, making it easier to cancel common components from the numerator and the denominator.
To factor a quadratic polynomial like our initial numerator, follow these steps:

• Look for two numbers that multiply to the product of the leading coefficient and the constant term and add up to the middle coefficient.
• Rewrite and split the middle term using these two numbers.
• Factor by grouping, aiming to express the polynomial as a product of binomials.

The given numerator, \(2x^2 + x - 6\), is factored into \((2x - 3)(x + 2)\). Similarly, the denominator \(x^2 + 3x + 2\) simplifies to \((x + 1)(x + 2)\). By factoring, the function can be reduced for further steps.

Domain of a Function

The domain of a function refers to all possible inputs (x-values) for which the function is defined.
For rational functions, the domain excludes values that cause the denominator to be zero since division by zero is undefined.
Start by setting the original denominator equal to zero and solve for \(x\):

• Solve \(x^2 + 3x + 2 = 0\) to find the x-values that make the denominator zero.
• The solutions are \(x = -1\) and \(x = -2\).

Thus, the domain of the original function, \(f(x)\), is all real numbers except \(x = -1\) and \(x = -2\). Identifying the domain is crucial for understanding where the function can be applied without issues.

Discontinuities in Functions

Discontinuities in functions occur at points where a function is not continuous. These include holes (removable discontinuities) and vertical asymptotes (non-removable discontinuities).
Both types of discontinuities arise from factors in the denominator of a rational function.

• A removable discontinuity, or a hole, occurs when a factor cancels out in the numerator and the denominator. In our function, \(x + 2\) cancels out, resulting in a hole at \(x = -2\).
• A vertical asymptote appears when a factor remains in the denominator after simplification. For the simplified function, \(x = -1\) creates a vertical asymptote.

Recognizing these discontinuities helps us accurately sketch the function's graph and understand its behavior near these critical points.

Asymptotes

Asymptotes are lines that a graph approaches but never actually touches. They are important in analyzing the behavior of rational functions as x-values approach certain limits.
There are two types relevant here: vertical and horizontal asymptotes.

• Vertical Asymptotes occur where the denominator is zero in the simplified function. In our example, this is at \(x = -1\).
• Horizontal Asymptotes depend on the degrees of the numerator and denominator. If these degrees are equal, the horizontal asymptote lies at \(y =\) the ratio of leading coefficients. In our case, \(y = 2\).

Understanding asymptotes helps predict the function's end behavior and how it behaves near points of indeterminacy.

Graphing Functions

Graphing functions involves plotting key features to illustrate their behavior.
For rational functions, this includes the domain, discontinuities, asymptotes, intercepts, and end behavior.

• Identify and plot any intercepts. Here, the x-intercept is where \(2x - 3 = 0\), or \(x = \frac{3}{2}\).
• Draw vertical asymptotes as dashed lines at \(x = -1\).
• Include a horizontal asymptote, \(y = 2\), as a dashed horizontal line.
• Mark any removable discontinuities (holes), such as at \(x = -2\).

Draw the curve through these points, respecting the asymptotes by ensuring the graph approaches them without touching at these lines. Seeing the graph helps to visualize all these properties and interpret the function's overall behavior.