Question

Sketch the graph of the equation. Use intercepts, extrema, and asymptotes as sketching aids. \(g(x)=\frac{x^{2}-x-2}{x-2}\)

Short answer

The graph for the function \(g(x)=\frac{x^{2}-x-2}{x-2}\) is a straight line with the equation \(y = x + 1\) with a vertical asymptote at \(x = 2\). It has x-intercept at \(-1\) and y-intercept at \(1\). The function does not have any minimum or maximum point.

Step-by-step solution

Simplification of the Function

The function in the problem can be factorized and simplified. The numerator of the fraction can be factorized to \(x^{2}-x-2 = (x-2)(x+1)\). Therefore, \(g(x)=\frac{(x-2)(x+1)}{x-2}\) and by cancelling out the common factors, we get simplified \(g(x)=x+1\). Note: This is under the condition \(x≠2\) because the denominator of the original function cannot be zero.

Find the Intercepts

Intercepts are points at which the graph crosses the axes:- The x-intercept is found by setting \(y = 0\) or \(g(x) = 0\). For this function \(0 = x+1\) thus \(x = -1\).- The y-intercept is found by setting \(x = 0\). For this function, \(g(x)=0+1=1\). So, the y-intercept is at \(y=1\).

Find the Extrema

The extrema points are found by taking the derivative of the function and equating it to zero. However, function \(g(x)=x+1\) is a straight line, hence it doesn't have a maximum or minimum point.

Find the Asymptotes

For the function \(g(x)=x+1\), there are no horizontal asymptotes since the function does not approach a specific value as \(x\) tends to positive or negative infinity. However, since \(x≠2\) in the original function, there is a vertical asymptote at \(x=2\).

Key concepts

Intercepts

Understanding intercepts is essential when graphing rational functions like the one given by the expression \(g(x) = \frac{x^2-x-2}{x-2}\). Intercepts can provide us with several crucial points where the graph intersects the axes.

In this context, there are two types of intercepts to determine: the x-intercept and the y-intercept. The x-intercept is where the graph crosses the x-axis. It's found by setting the output of the function, commonly represented as \(y\), to zero. In our case, setting \(g(x) = 0\) results in the equation \(0 = x + 1\), leading us to the single x-intercept at \(x = -1\).

The y-intercept, on the other hand, occurs where the graph crosses the y-axis, at which point \(x = 0\). Substituting \(x = 0\) into the function provides us with \(g(x) = 0 + 1 = 1\), positioning the y-intercept at \(y = 1\). This gives us a clear starting point to sketch our graph with the confidence that it passes through the points \((-1, 0)\) and \((0, 1)\).

Extrema

When we refer to 'extrema' in the realm of mathematics, we are typically talking about the highest or lowest points on a graph; these are known as maxima and minima respectively. To seek out extrema, one would usually calculate the derivative of the function and determine where this derivative is equal to zero, indicating a potential maximum or minimum.

However, the rational function provided simplifies to a linear function, \(g(x) = x + 1\), after factorization and cancellation. Linear functions, being straight lines, do not have the hills and valleys found in curves; hence, they do not possess maxima or minima. Thus, despite the importance of extrema in analyzing functions overall, our particular function does not offer extrema—a noteworthy realization that informs us about the function's behavior and aids in sketching its graph.

Vertical Asymptotes

A vertical asymptote in graphing rational functions is a vertical line which the graph approaches but never touches or crosses. It represents values that the input, or \(x\), cannot be equal to, due to restrictions like division by zero.

In the original function \(g(x) = \frac{x^2-x-2}{x-2}\), we factor and simplify, but we must recall that we did so under the condition that \(x eq 2\), since the denominator of the original function cannot be zero. Respectfully, even though the simplified version appears to be the line \(g(x) = x + 1\), we remember that at \(x = 2\) there would've been a division by zero in the original equation. As a result, there exists a vertical asymptote at \(x = 2\).

While the line \(x = 2\) is not apparent in our simplified equation, its presence is fundamental to the graph of the original rational function and dictates that the graph will draw infinitely close to, but will never intersect or reach, the line \(x = 2\). When sketching the graph, it is crucial to include this asymptote to properly represent the function's behavior near this excluded value.