Problem 41 - Precalculus: Functions and Graphs

Question

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

Step-by-Step Solution

Answer

The simplified function is \(f(x) = x - 1\) with a hole at \(x = -2\). The graph is a line with a hole.

Step 1: Factor the numerator

First, we need to factor the quadratic expression in the numerator of the function. The expression is \(x^2 + x - 2\). We are looking for two numbers that multiply to \(-2\) and add to \(1\). These numbers are \(2\) and \(-1\). Therefore, \(x^2 + x - 2\) can be factored as \((x + 2)(x - 1)\).

Step 2: Simplify the expression

Now we have the expression \(f(x) = \frac{(x+2)(x-1)}{x+2}\). We can cancel the common factor \((x+2)\) in the numerator and the denominator, keeping in mind that \(x eq -2\) to avoid division by zero. The simplified form of \(f(x)\) is \(f(x) = x - 1\), except at \(x = -2\).

Step 3: Identify the hole in the graph

Since we cancelled \(x+2\) from both the numerator and the denominator, there is a hole in the graph at \(x = -2\). This is where the function is not defined because the original denominator was zero at this point.

Step 4: Sketch the graph of the function

The simplified function \(f(x) = x - 1\) is a straight line with a slope of \(1\) and a y-intercept at \(-1\). To sketch the graph, draw a line that intercepts the y-axis at \(-1\) and goes through the points such as \((0,-1)\), \((1,0)\), etc. Remember to indicate that there is a hole at \(x = -2\), which is represented by an open circle at the point \((-2, -3)\) on the graph.

Key Concepts

Factoring Polynomials

Factoring polynomials is a fundamental technique that involves breaking down a polynomial expression into a product of simpler polynomials. It is similar to factoring numbers but focuses on expressions with variables. In this exercise, we needed to factor the numerator of the given function, which was a quadratic polynomial: \(x^2 + x - 2\). We looked for two numbers that multiply to the constant term, \(-2\), and add up to the coefficient of the linear term, \(1\). Here, the numbers were \(2\) and \(-1\). Thus, the quadratic expression factors into \((x + 2)(x - 1)\).
Understanding how to factor polynomials is crucial in simplifying expressions and solving equations. It often serves as a critical step in many problems, especially those involving rational functions.

Function Simplification

Function simplification involves reducing a complex expression to its most basic form. For the exercise, we achieved this by first factoring the polynomial in the numerator as \((x+2)(x-1)\) and then canceling the common factor \((x+2)\) from both the numerator and the denominator. This resulted in the simplified function \(f(x) = x - 1\).
It's important to remember the conditions for which the simplification is valid. In this case, we couldn't cancel \(x+2\) unless we excluded \(x = -2\), since division by zero is undefined. Simplifying functions helps in making the expression easier to evaluate and graph, improving our ability to understand and interpret the behavior of functions.

Graphing Linear Functions

Graphing linear functions is a straightforward process once the function is in its simplified linear form, such as \(f(x) = x - 1\). This is a simple linear equation with:

• A slope of \(1\) (which means for every unit increase in \(x\), \(y\) increases by the same amount),
• A y-intercept of \(-1\) (where the line crosses the y-axis).

To sketch the graph, we plot the y-intercept, then use the slope to find other points by moving one unit to the right and one unit up. For example, starting at the y-intercept \((0, -1)\), the next point would be \((1, 0)\). Drawing a line through these points gives the graph of the function.
Understanding how to graph linear functions is essential, as it visually represents how the function behaves, making it easier to analyze and interpret.

Holes in Graphs

Holes in graphs occur in rational functions when a factor is canceled from both the numerator and the denominator, leaving a point undefined. In our exercise, this happened at \(x = -2\), where the factor \((x+2)\) was canceled. This results in a hole, represented graphically as an open circle on the plot.
It's crucial to consider these holes since they mark points where the function does not exist, even though the simplified graph continues through them. For our function, the hole is located at the point \((-2, -3)\), which originally would be a valid point if not for the cancellation. Recognizing and marking holes helps avoid misinterpretations of the function's behavior and ensures analytical calculations remain accurate and meaningful.