Problem 59 - 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. $$f(x)=-\left(x^{2}-1\right)(x-2)(x+3)$$

Step-by-Step Solution

Answer

The function \(f(x)= -x^{2}\) has the same end behavior as \(f(x)=-\left(x^{2}-1\right)(x-2)(x+3)\). The x-intercepts are \(-3\), \(-1\), and \(2\), and the y-intercept is \(-6\). The function is positive for \(x < -3\) and \(x > 2\), and negative for \(-3 < x < -1\) and \(-1 < x < 2\).

Step 1: Determine the end behavior

The degree of the polynomial is 2 + 1 + 1 = 4, which is even, and the leading coefficient is negative. Therefore, as x goes to positive or negative infinity, \(f(x)\) approaches positive infinity. A quadratic function with the same end behavior is \(y=-x^{2}\).

Step 2: Find the x- and y-intercepts

The x-intercepts can be found by setting \(f(x)\) equal to zero and solving for x: \(-\left(x^{2}-1\right)(x-2)(x+3)=0\). This leads to \(x=\pm1, 2, -3\). The y-intercept can be found by setting x equal to zero in \(f(x)\), which gives \(f(0)= -\left(0^{2}-1\right)(0-2)(0+3) = -6\)

Step 3: Determine the intervals where the function is positive and negative

Knowing the x-intercepts, we can examine the intervals of \(x\) between and beyond these points. Choose test points in these intervals and substitute into \(f(x)\): For \(x < -3\), choose \(x= -4\), then \(f(-4) > 0\). For \(-3 < x < -1\), choose \(x= -2\), then \(f(-2) < 0\). For \(-1 < x < 2\), choose \(x= 0\), then \(f(0) < 0\). For \(x > 2\), choose \(x= 3\), then \(f(3) > 0\). Therefore, \(f(x)\) is positive for \(x < -3\) and \(x > 2\), and negative for \(-3 < x < -1\) and \(-1 < x < 2\).

Step 4: Drawing the graph of \(f(x)\)

Plot the x and y intercepts onto a graph. The intercepts at x= -3, -1, 2 and y= -6. Sketch a curve passing through these points that reflects the determined end behavior and regions of positivity and negativity. The curve starts and ends up high (since \(f(x)\) approaches positive infinity as \(x\) approaches both positive and negative infinity), dips below the x-axis between \(x = -1\) and \(x = 2\) , and rises above the x-axis when \(x < -3\) or \(x > 2\) .

Key Concepts

End Behavior of Polynomials

The end behavior of a polynomial function describes how the function behaves as the input values, or x, approach positive or negative infinity. This behavior is largely determined by two main components: the degree of the polynomial and its leading coefficient.

For our given polynomial, the degree is 4, calculated as the sum of the exponents: 2, 1, and 1. This is an even degree, and since the leading coefficient is negative, the function's behavior is as follows:

• As the input values approach positive infinity (\(x \to \infty\)), the value of the function (\(f(x)\)) approaches positive infinity.
• Similarly, as the input values approach negative infinity (\(x \to -\infty\)), the value of the function remains consistently approaching positive infinity.

To match this behavior with a simpler quadratic function, we use \(y = -x^{2}\), exemplifying similar end behavior with an upside-down parabolic shape.

Graph Intercepts

Graph intercepts are the points where a polynomial function crosses the x-axis and y-axis. These points help in understanding the shape and position of the graph relative to the axes.

**X-intercepts:**
To find the x-intercepts, we solve the equation \(f(x) = 0\). In this case, the factored form of the polynomial gives us:

• \(x^{2} - 1 = 0\), leading to \(x = \pm 1\).
• \(x - 2 = 0\), resulting in \(x = 2\).
• \(x + 3 = 0\), giving \(x = -3\).

**Y-intercept:**
The y-intercept is found by setting x to 0 in the function. Substituting gives: \(f(0) = -((-1)(-6)) = -6\). Therefore, the y-intercept is at \(y = -6\). These intercepts are crucial for plotting key points on the graph and envisioning its progression.

Intervals of Positivity and Negativity

Understanding where a polynomial is positive or negative involves examining the function's sign in different segments along the x-axis. The x-intercepts help define these segments as they divide the x-axis into intervals.

For the given polynomial, the intervals formed are:

• \(x < -3\): Positive. Testing with \(x = -4\) shows \(f(-4) > 0\).
• \(-3 < x < -1\): Negative. Testing with \(x = -2\) gives \(f(-2) < 0\).
• \(-1 < x < 2\): Negative. Testing with \(x = 0\) provides \(f(0) < 0\).
• \(x > 2\): Positive. Testing with \(x = 3\) results in \(f(3) > 0\).

These findings show where the polynomial lies above or below the x-axis, completing the picture needed to sketch the graph accurately. By determining these intervals, you can predict how the function behaves between and beyond its intercepts, aiding in constructing a reliable visual representation.