Problem 52 - Precalculus : Building Concepts and Connections

Question

For each polynomial function, (a) find a function of the form \(y=c x^{2}\) that has the same end behavior. (b) find the \(x\) - and \(y\) -intercept(s) of the graph. (c) find the interval(s) on which the value of the function is positive. (d) find the interval(s) on which the value of the function is negative. (e) use the information in parts ( \(a\) ) \(-\) (d) to sketch a graph of the function. $$f(x)=(x+1)(x-2)(x+3)$$

Step-by-Step Solution

Answer

a) \(y=x^3\), b) \(x=-1,2,-3\) and \(y=6\), c) \((-3,-1)\) and \((2,+\infty)\), d) \((-\infty,-3)\) and \((-1,2)\). The graph is sketched according to these findings.

Step 1: Finding the end behavior

To determine the end behavior of the function \(f(x) = (x+1)(x-2)(x+3)\), expand the equation to obtain \(f(x) = x^3 + 2x^2 - 5x - 6\). As \(x\) approaches positive or negative infinity, \(x^3\) will dominate. Therefore, the end behavior of the function will be similar to the function \(y = x^3\). The end behavior of \(y = x^3\) is \(y = -\infty\) as \(x \rightarrow -\infty\) and \(y = +\infty\) as \(x \rightarrow +\infty\). In the form \(y = cx^2\), this is \(y = cx^3\), where \(c = 1\).

Step 2: Finding the x and y intercepts

The x-intercepts of the function are the solutions to the equation \(f(x) = 0\). Setting each factor of \(f(x) = (x + 1)(x - 2)(x + 3)\) equal to zero gives \(x = -1, 2, -3\). These are the x-intercepts. The y-intercept is obtained by setting \(x = 0\). Substituting \(x = 0\) into \(f(x)\) gives \(f(0) = (-1)(-2)(3) = 6\). Therefore, the y-intercept is (0, 6).

Step 3: Determining where the function is positive or negative

To determine where the function is positive or negative, one can examine the sign of the function for values on each interval between the x-intercepts. Choosing test points -2, 0, and 3 to represent the intervals \((- \infty, -3)\), \((-3, -1)\), \((-1, 2)\) and \((2, \infty)\) respectively, we find that f(-2) = -20, f(0) = 6, f(1) = 4, and f(3) = 40. The function is therefore negative on the intervals \((- \infty, -3)\) and \((-1, 2)\), and positive on the intervals \((-3, -1)\) and \((2, \infty)\).

Step 4: Sketching the graph

Finally, sketch the graph. Plot the x and y intercepts first. The graph should pass through these points. Then, consider the intervals on which the function is positive or negative, and use this information to sketch the overall shape of the graph. Starting from \(-\infty\), the graph is negative until \(x = -3\), then positive until \(x = -1\), negative until \(x = 2\), and finally positive as \(x\) approaches \(+\infty\).

Key Concepts

End Behavior

Understanding the end behavior of a polynomial function helps us predict how the function behaves as the input values become extremely large or extremely small. For the polynomial function given, \[ f(x) = (x+1)(x-2)(x+3), \]expanding it results in \[ f(x) = x^3 + 2x^2 - 5x - 6. \]Here, the term with the highest degree, which is \( x^3 \), dictates the end behavior. As \( x \to \infty \), \( f(x) \to \infty \) because the \( x^3 \) term dominates and grows positively larger. Similarly, as \( x \to -\infty \), \( f(x) \to -\infty \). This tells us that the end behavior resembles that of a simple cubic function:

• \( y = -\infty \) as \( x \to -\infty \)
• \( y = +\infty \) as \( x \to +\infty \)

This predicts how the function behaves at the outer extremes, assisting in the sketching process.

Intercepts

Intercepts are crucial for understanding where the graph crosses the axes. The polynomial \[ f(x) = (x+1)(x-2)(x+3), \]has x-intercepts where each individual factor equals zero. Solving

• \( x + 1 = 0 \) gives \( x = -1 \)
• \( x - 2 = 0 \) gives \( x = 2 \)
• \( x + 3 = 0 \) gives \( x = -3 \)

Thus, the x-intercepts are points where the graph crosses the x-axis: \((-1, 0), (2, 0), \text{ and } (-3, 0)\). Calculating the y-intercept involves setting \( x = 0 \), which yields:\[ f(0) = (0+1)(0-2)(0+3) = 6. \]So, the y-intercept is at \((0, 6)\). These intercepts provide important anchor points for sketching the curve.

Intervals of Positivity and Negativity

Determining where a polynomial function is positive or negative involves assessing the sign of the function across different intervals. Using the x-intercepts \(-3, -1, \text{ and } 2\) to separate the number line into segments, test points can be chosen from each interval to identify the behavior:

• Interval \((-\infty, -3)\): Pick \( x = -4 \), \( f(-4) < 0 \). Function is negative.
• Interval \((-3, -1)\): Pick \( x = -2 \), \( f(-2) > 0 \). Function is positive.
• Interval \((-1, 2)\): Pick \( x = 0 \), \( f(0) > 0 \). Function is positive.
• Interval \((2, \infty)\): Pick \( x = 3 \), \( f(3) > 0 \). Function is positive.

This analysis reveals that the function \( f(x) \) is negative on \((-\infty, -3)\) and \((-1, 2)\), and positive on \((-3, -1)\) and \((2, \infty)\). This information about where the function sits relative to the x-axis aids in understanding and sketching the graph.

Graph Sketching

Graph sketching is the culmination of understanding a function's behavior. Start by plotting the intercepts: the x-intercepts at \((-3, 0), (-1, 0), \text{ and } (2, 0)\) and the y-intercept at \((0, 6)\). Then, consider the intervals of positivity and negativity:

• The graph falls below the x-axis from \(-\infty\) to \(x = -3\).
• It rises above the x-axis between \(-3\) and \(-1\).
• Between \(-1\) and 2, it stays above the x-axis.
• Finally, from \(x = 2\) onwards, the graph is again above the x-axis, moving upwards indefinitely.

The transitions at each root (where the function crosses the x-axis) are crucial. The graph will exhibit a cubic 'S' shape due to the relationship of the intercepts and the nature of the cubic function. Combining these insights, you can effectively sketch an accurate representation of the polynomial function.