Problem 17 - Precalculus Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

The x-intercepts are -2, 0, and 3; the end behavior extends from negative infinity to positive infinity.

Step 1: Identify the degree and leading coefficient

The function \(P(x) = x(x-3)(x+2)\) is a polynomial. When expanded, it has three factors: \(x\), \((x-3)\), and \((x+2)\). Multiplying these, we have \(x^3 - x^2 - 6x\). Thus, it is a cubic polynomial. The degree of the polynomial is 3, and the leading coefficient is 1.

Step 2: Determine End Behavior

Since the degree of the polynomial is odd and the leading coefficient is positive, the end behavior of the polynomial is such that as \(x o -\infty\), \(P(x) o -\infty\) and as \(x o +\infty\), \(P(x) o +\infty\).

Step 3: Find the x-intercepts

Set \(P(x) = 0\) and solve: \(x(x-3)(x+2) = 0\). This gives the x-intercepts of the polynomial: \(x = 0\), \(x = 3\), and \(x = -2\). The polynomial crosses the x-axis at these points.

Step 4: Find the y-intercept

To find the y-intercept, substitute \(x = 0\) into \(P(x)\): \(P(0) = 0(0-3)(0+2) = 0\). Thus, the y-intercept is \( (0, 0) \).

Step 5: Sketch the graph based on intercepts and end behavior

Plot the x-intercepts \((0, 0)\), \((3, 0)\), and \((-2, 0)\), and the y-intercept \((0, 0)\). Using the determined end behavior, start the graph from below the x-axis for \(x < -2\), cross the x-axis at \((-2,0)\), ascend, pass through the origin \((0,0)\), decrease to cross again at \((3,0)\), and then continue upwards.

Key Concepts

Understanding x-intercepts in a Polynomial Function

The x-intercepts are the points where the graph of a polynomial crosses the x-axis. For the polynomial function \(P(x) = x(x-3)(x+2)\), finding the x-intercepts involves setting the polynomial equal to zero and solving for \(x\). This is because, at these points, the output of the function (or \(y\)-value) is zero.
To find the x-intercepts for \(P(x)\), we set:\[x(x-3)(x+2) = 0\]Now, solve for \(x\):

• \(x = 0\)
• \(x - 3 = 0\) leads to \(x = 3\)
• \(x + 2 = 0\) leads to \(x = -2\)

Therefore, the x-intercepts are \((-2, 0)\), \((0, 0)\), and \((3, 0)\).
By analyzing the function’s equation, you can visually anticipate where the graph interacts with the x-axis. These points are crucial for sketching the overall graph of the polynomial.

Exploring the End Behavior of Polynomial Functions

The end behavior of a polynomial function describes how the graph of the function behaves as \(x\) approaches positive or negative infinity. For \(P(x) = x(x-3)(x+2)\), determining the end behavior helps us understand what happens to the graph at the extreme ends.
Since this is a cubic function with an odd degree (3) and a positive leading coefficient (1), the end behavior demonstrates a specific pattern:

• As \(x\) approaches \(-\infty\) (very large negative values), \(P(x)\) approaches \(-\infty\).
• As \(x\) approaches \(+\infty\) (very large positive values), \(P(x)\) approaches \(+\infty\).

This pattern means the graph begins low (negative y-values) on the left, rises and crosses the x-axis at the intercepts, and finishes high (positive y-values) on the right. This end behavior is critical for appropriately sketching the polynomial, as it sets the trajectory of the graph beyond the visible intercepts.

Determining the y-intercept

The y-intercept of a polynomial is where the graph crosses the y-axis. At this point, the x-value is zero. For the function \(P(x) = x(x-3)(x+2)\), to find the y-intercept, we substitute \(x = 0\) into the equation of the polynomial:
\[P(0) = 0(0-3)(0+2) = 0\]
Hence, the y-intercept is at the point \((0, 0)\). This means the graph of the polynomial passes through the origin of the coordinate plane.
The y-intercept provides a clear starting point or reference for drawing the graph and shows how the polynomial behaves as it crosses the y-axis. Remember, the y-intercept is always a single point unless the graph bounces or touches but doesn’t cross, common in higher multiplicity factors which isn’t the case in this example.