Problem 13 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The x-intercepts are at x = 0, 3, -2; y-intercept at (0,0); end behavior: up to the right, down to the left.

Step 1: Identify the degree of the polynomial

The polynomial function is given by \( P(x) = x(x-3)(x+2) \). To find the degree, we count the factors of \( x \) in the expression. Multiplying the factors gives us \( x (x-3) (x+2) = x^3 - x^2 - 6x \), which is a polynomial of degree 3 because the highest power of \( x \) is 3.

Step 2: Find the x-intercepts

To find the x-intercepts, set the polynomial equal to zero: \( x(x-3)(x+2) = 0 \). Solving this equation gives the x-intercepts at \( x = 0 \), \( x = 3 \), and \( x = -2 \). These are the values where the graph intersects the x-axis.

Step 3: Determine the end behavior

For polynomial functions, the end behavior is determined by the leading term (the term with the highest degree). Here, the leading term is \( x^3 \). Since its coefficient is positive, as \( x \rightarrow +\infty \), \( P(x) \rightarrow +\infty \) and as \( x \rightarrow -\infty \), \( P(x) \rightarrow -\infty \).

Step 4: Evaluate the y-intercept

To find the y-intercept, evaluate \( P(0) \). Plugging in \( x = 0 \) gives \( P(0) = 0(0-3)(0+2) = 0 \). So, the y-intercept is at the origin, (0, 0).

Step 5: Sketch the graph

Begin by plotting the x-intercepts, (0, 0), (3, 0), and (-2, 0), and the y-intercept, (0, 0). Then, incorporate the end behavior: as \( x \rightarrow +\infty \), \( P(x) \rightarrow +\infty \) and as \( x \rightarrow -\infty \), \( P(x) \rightarrow -\infty \). Draw a smooth curve passing through the intercepts that reflects this behavior.

Key Concepts

Degree of Polynomial

The degree of a polynomial is crucial in understanding the overall shape of its graph. In the polynomial function given as \( P(x) = x(x-3)(x+2) \), the degree is determined by the highest power of \( x \) when the expression is fully expanded. Let's break it down: multiplying out the terms, we have \( x (x-3) (x+2) = x^3 - x^2 - 6x \).

This results in the highest power of \( x \) being 3, which means the function is a cubic polynomial. A degree of 3 often indicates the graph will cross the x-axis at three different points and exhibit a particular cubic-like shape, featuring either one hill and one valley or potentially more intricate curves depending on additional polynomial terms.

X-intercepts

X-intercepts are the points where the graph meets the x-axis. These are fundamental for sketching the graph and understanding the behavior of the function. To find them in the polynomial \( P(x) = x(x-3)(x+2) \), we set the polynomial function equal to zero:

\( x(x-3)(x+2) = 0 \).

We solve for \( x \) by setting each factor of the polynomial to zero individually:

• From \( x = 0 \), we have an intercept at (0, 0).
• From \( x-3 = 0 \), we get \( x = 3 \), which means there is an intercept at (3, 0).
• From \( x+2 = 0 \), we find \( x = -2 \), providing an intercept at (-2, 0).

These intercepts are the points where the polynomial touches or crosses the x-axis, informing the placement of key points when sketching the graph.

End Behavior

The end behavior of a polynomial function reveals how the graph acts as \( x \) approaches positive or negative infinity. This is governed by the leading term, which, for the polynomial \( P(x) = x^3 - x^2 - 6x \), is \( x^3 \).

Since the leading coefficient of \( x^3 \) is positive, the graph will behave in the following manner:

• As \( x \rightarrow +fty \), \( P(x) \rightarrow +fty \).
• As \( x \rightarrow -fty \), \( P(x) \rightarrow -fty \).

This indicates that the graph rises on the right side and falls on the left side, typical of an odd degree polynomial with a positive leading coefficient. Understanding this behavior is essential for correctly sketching the overall graph.

Y-intercept

The y-intercept of a polynomial function tells us where the graph crosses the y-axis. It occurs when \( x = 0 \). For the function \( P(x) = x(x-3)(x+2) \), we calculate the y-intercept by evaluating the function at \( x = 0 \):

\( P(0) = 0(0-3)(0+2) = 0 \).

Thus, the y-intercept is located at (0, 0).

In this particular case, it is noteworthy that the y-intercept coincides with one of the x-intercepts, indicating that the graph crosses both the x and y-axes at the origin. This point is important as it confirms that the origin is a critical point where the function intersects both axes.