Problem 19 - Algebra and Trigonometry

Question

Sketch the graph of the polynomial function. Make sure your graph shows all intercepts and exhibits the proper end behavior. $$ P(x)=\frac{1}{12}(x+2)^{2}(x-3)^{2} $$

Step-by-Step Solution

Answer

The graph of \(P(x)\) rises on both ends, touches and rebounds at \((-2,0)\) and \((3,0)\), with a y-intercept at \((0,3)\).

Step 1: Identify the Type of Function

We start with the polynomial function \(P(x) = \frac{1}{12}(x+2)^{2}(x-3)^{2}\). This is a fourth-degree polynomial function because both factors \((x+2)\) and \((x-3)\) are squared, resulting in an \(x^4\) term when expanded.

Step 2: Determine the Zeros and Their Multiplicities

The zeros of \(P(x)\) are the solutions to \((x+2) = 0\) and \((x-3) = 0\). Thus, the roots are \(x = -2\) and \(x = 3\), each with multiplicity 2, indicating the graph touches the x-axis at these points and bounces off rather than cross it.

Step 3: Calculate the Y-intercept

Set \(x = 0\) to find the y-intercept. Substitute into \(P(x)\): \[P(0) = \frac{1}{12}(0+2)^2(0-3)^2 = \frac{1}{12} \times 4 \times 9 = 3.\]Hence, the y-intercept is \((0, 3)\).

Step 4: Analyze the End Behavior

Since the polynomial is of fourth-degree and the leading coefficient is positive, both ends of the graph will rise to positive infinity as \(x\) goes to positive or negative infinity. Thus, as \(x \rightarrow -\infty\), \(P(x) \rightarrow +\infty\), and as \(x \rightarrow +\infty\), \(P(x) \rightarrow +\infty\).

Step 5: Sketch the Graph

Using the information from the previous steps, sketch the graph:1. Note the intercepts at \((-2, 0)\), \((3, 0)\), and \((0, 3)\).2. The graph touches the x-axis and rebounds at the points \((-2, 0)\) and \((3, 0)\).3. The graph rises on both sides because of the end behavior.Draw a smooth curve that exhibits these characteristics.

Key Concepts

Zeros and Multiplicity

In polynomial functions, zeros are the values of \(x\) where the function equals zero. For the polynomial \(P(x) = \frac{1}{12}(x+2)^{2}(x-3)^{2}\), the zeros are found by setting each factor within the expression equal to zero. Thus, the zeros are \(x = -2\) and \(x = 3\). These zeros have a multiplicity of 2, because each zero comes from a factor that is squared. Multiplicity tells us how the graph behaves at the intercept. If the multiplicity is odd, the graph crosses the x-axis. However, for an even multiplicity like 2, the graph touches the x-axis at the zero and then "bounces" back. At \(x = -2\) and \(x = 3\), the graph will not cut through the x-axis but will instead touch it and turn back.

End Behavior

The end behavior of a polynomial function is how the graph behaves as \(x\) approaches positive or negative infinity. For our function, the polynomial \(P(x) = \frac{1}{12}(x+2)^{2}(x-3)^{2}\) is a fourth-degree polynomial, due to its highest power being 4 when expanded. The leading coefficient, which is the constant that comes before the term with the highest power, is positive here (\(\frac{1}{12}\)). For any polynomial where the leading term has a degree that is even, and if its leading coefficient is positive, both ends of the polynomial will rise toward positive infinity. This means as \(x\) moves to negative infinity, \(P(x)\) also increases towards infinity, and similarly as \(x\) moves to positive infinity.

Y-intercept

The y-intercept is the point where the graph crosses the y-axis. This can be found by evaluating the polynomial function at \(x = 0\). For \(P(x) = \frac{1}{12}(x+2)^{2}(x-3)^{2}\), setting \(x\) to 0 gives:

• \(P(0) = \frac{1}{12}(0+2)^2(0-3)^2 = \frac{1}{12} \times 4 \times 9 = 3\).

The y-intercept is determined to be \((0, 3)\). This means the graph of the polynomial will cross the y-axis at the point where \(y = 3\). Understanding the y-intercept helps us to pin down at least one point on the graph, aiding in accurately sketching the curve.

Fourth-degree Polynomial

A fourth-degree polynomial means that the highest power of \(x\) in the polynomial is 4. For the function \(P(x) = \frac{1}{12}(x+2)^{2}(x-3)^{2}\), expanding the expression results in a polynomial of the form \(ax^4 + bx^3 + cx^2 + dx + e\), where the leading term has the power of 4. Such polynomials tend to have certain characteristics:

• They can have up to 4 real roots (though not necessarily all unique).
• They can cross or touch the x-axis up to 4 times based on the multiplicities of these roots.
• Their graphs typically appear more undulating or smooth as compared to lower-degree polynomials.

For our polynomial, knowing it is a fourth degree with a positive leading coefficient gives insight into the general shape and direction of the graph, particularly its end behavior.