Problem 12 - Algebra and Trigonometry

Question

Sketch the graph of the polynomial function. Make sure your graph shows all intercepts and exhibits the proper end behavior. $$ P(x)=(x-1)(x+1)(x-2) $$

Step-by-Step Solution

Answer

The x-intercepts are at x = -1, 1, 2, the y-intercept is at (0, 2), with a cubic end behavior.

Step 1: Determine the x-intercepts

The x-intercepts of the polynomial function \( P(x) = (x-1)(x+1)(x-2) \) are the solutions to the equation \( P(x) = 0 \). Each factor provides an x-intercept: \( x-1 = 0 \) gives \( x = 1 \), \( x+1 = 0 \) gives \( x = -1 \), and \( x-2 = 0 \) gives \( x = 2 \). So, the x-intercepts are \( x = -1, 1, \) and \( 2 \).

Step 2: Determine the y-intercept

To find the y-intercept, evaluate \( P(x) \) at \( x = 0 \). \( P(0) = (0 - 1)(0 + 1)(0 - 2) = (-1)(1)(-2) = 2 \). Thus, the y-intercept is \( (0, 2) \).

Step 3: Determine end behavior

The end behavior of a polynomial function is determined by the leading term when the function is expanded. For \( P(x) = (x-1)(x+1)(x-2) \), the leading term is \( x^3 \). Since \( x^3 \) has an odd degree and a positive leading coefficient, \( P(x) \rightarrow -\infty \) as \( x \rightarrow -\infty \) and \( P(x) \rightarrow \infty \) as \( x \rightarrow \infty \).

Step 4: Sketch the graph

Now plot the intercepts on a graph. The graph will pass through the x-intercepts at \( (-1,0) \), \( (1,0) \), and \( (2,0) \), and the y-intercept at \( (0,2) \). Considering the end behavior, draw the curve such that as it moves left of \( x = -1 \), it dips down to negative infinity, and as it moves right of \( x = 2 \), it heads up to positive infinity. The curve will cross the x-axis at each intercept and respect the y-intercept.

Key Concepts

Understanding x-intercepts of a polynomial function

X-intercepts are the points where a graph crosses the x-axis. These are also known as the roots or solutions to the equation when the function equals zero. In our polynomial function, \( P(x) = (x-1)(x+1)(x-2) \), finding the x-intercepts involves setting each factor equal to zero and solving the equations.

• For \( x - 1 = 0 \), we determine that \( x = 1 \).
• For \( x + 1 = 0 \), we get \( x = -1 \).
• For \( x - 2 = 0 \), the solution is \( x = 2 \).

Each of these solutions tells us where the graph will intersect the x-axis, specifically at points \(-1, 0\), \(1, 0\), and \(2, 0\). These points are significant because they help us plot the graph accurately.

Determining the y-intercept

The y-intercept of a polynomial function is where the graph crosses the y-axis. It is found by evaluating the polynomial at \( x = 0 \). For the function \( P(x) = (x-1)(x+1)(x-2) \):

• Substitute \( x = 0 \) into the polynomial: \( P(0) = (0 - 1)(0 + 1)(0 - 2) \).
• This results in \( P(0) = (-1)(1)(-2) = 2 \).

Hence, the y-intercept is located at the point \( (0, 2) \). This point is crucial for graph sketching, as it provides a definitive point where the graph will cross the vertical y-axis.

Exploring the end behavior of polynomial functions

End behavior describes what happens to the values of a function as \( x \) moves towards infinity or negative infinity. It is particularly influenced by the degree and leading coefficient of the polynomial. For \( P(x) = (x-1)(x+1)(x-2) \), when expanded, the leading term is \( x^3 \):

• The degree is 3, which is odd.
• The leading coefficient is positive (1).

This combination implies a specific pattern:

• As \( x \to -\infty \), \( P(x) \to -\infty \) because the graph extends downward to the left.
• As \( x \to \infty \), \( P(x) \to \infty \) because the graph extends upward to the right.

Understanding this behavior helps predict the overall shape of the graph beyond immediate plotted points.

Practical steps for graph sketching

Graph sketching is about visually representing the function based on key characteristics like intercepts and end behavior. For the polynomial \( P(x) = (x-1)(x+1)(x-2) \):First, plot the x-intercepts \((-1,0)\), \((1,0)\), and \((2,0)\) on your graph. These points show where your graph will cross the x-axis.
Next, plot the y-intercept \((0,2)\). This provides a starting point for the vertical placement of the graph.
After plotting these crucial points, consider the end behavior. From the end behavior analysis, we know:

• The graph dips down towards negative infinity as it moves left of \(x = -1\).
• The graph rises toward positive infinity as it moves right of \(x = 2\).

Connecting these points considering the end behavior, sketch a smooth curve through the intercepts.By accounting for all these aspects, you capture the correct shape and positioning of the graph on a coordinate plane.