Problem 55 - Algebra and Trigonometry

Question

In Exercises 41–64, a. Use the Leading Coefficient Test to determine the graph’s end behavior. b. Find the x-intercepts. State whether the graph crosses the x-axis, or touches the x-axis and turns around, at each intercept. c. Find the y-intercept. d. Determine whether the graph has y-axis symmetry, origin symmetry, or neither. e. If necessary, find a few additional points and graph the function. Use the maximum number of turning points to check whether it is drawn correctly. $$f(x)=-3(x-1)^{2}\left(x^{2}-4\right)$$

Step-by-Step Solution

Answer

The end behavior of the graph is: falls to the left and falls to the right. The x-intercepts are at x = 1 (touches and turns around), x = -2 and x = 2 (crosses at both). The y-intercept is at (0, -3). The graph has neither y-axis symmetry nor origin symmetry. On graphing, it should have at most 3 turning points.

Step 1: Determine End Behavior

To see the end behavior of the graph of a function, examine the leading coefficient and the degree of the function. The degree of the function \(f(x)=-3(x-1)^{2}(x^{2}-4)\) is 4 (adding up the powers of x), and the leading coefficient is -3. Since the degree is even and the leading coefficient is negative, according to the Leading Coefficient Test, the graph falls to the left and falls to the right.

Step 2: Find the x-intercepts

To find the x-intercepts, set the function \(f(x)\) equal to zero and solve for \(x\). So, -3(x-1)^{2}(x^{2}-4) = 0, resulting in x-intercepts at x = 1, x = -2, and x = 2. At x = 1, the graph touches the x-axis and turns around, because the power of the corresponding factor in the function is 2 (being even, the graph touches and turns around). For x = -2 and x = 2, the graph crosses the x-axis (being an odd power, the graph crosses at the intercept).

Step 3: Find the y-intercept

The y-intercept is found by setting \(x = 0\) in the function. So, substituting x with 0, \(f(x)\) equals -3, which gives the y-intercept as (0, -3).

Step 4: Check for Symmetry

To decide whether the graph has y-axis symmetry, origin symmetry, or neither, first replace \(x\) with \(-x\). If \(f(x) = f(-x)\), the graph has y-axis symmetry. If \(f(x) = -f(-x)\), the graph has origin symmetry. Avoiding confusion, when both conditions fail, the graph has neither symmetry. Upon substitution, it's seen that neither equation is true, implying that the graph has neither y-axis symmetry, nor origin symmetry.

Step 5: Graph the function

This step entails plotting a few additional points and sketching the function. The x-intercepts (1, -2, and 2), y-intercept (0, -3), the symmetry details, and the end behavior details acquired in previous steps will guide in plotting and shaping the graph accurately. It's also useful to know that a polynomial function could have at most \(n-1\) turning points, where \(n\) is the degree, in this case, 4-1=3. This also helps to confirm whether the sketch of the graph is correct.

Key Concepts

Leading Coefficient Test

The Leading Coefficient Test is a simple way to predict the end behavior of a polynomial function's graph, based on its leading coefficient and the degree of the polynomial.

Specifically, if a polynomial has an even degree and a positive leading coefficient, the graph will rise to the left and to the right. Conversely, an even degree with a negative leading coefficient means the graph falls in both directions. If the degree is odd, a positive leading coefficient results in the graph falling to the left and rising to the right, while a negative coefficient reverses that behavior, rising to the left and falling to the right. For the given function \(f(x)=-3(x-1)^{2}(x^{2}-4)\), with an even degree of 4 and a negative leading coefficient of -3, the graph falls on both ends.

X-intercepts of a function

The x-intercepts of a function are points where the graph crosses or touches the x-axis, which occur at values of x for which \(f(x)=0\). These points can reflect where a function changes direction, pointing to the function's behavior and indicating potential turning points.

To find the x-intercepts algebraically, we solve for x in the equation \(f(x)=0\). In the step-by-step solution, the x-intercepts are found to be at \(x=1\), \(x=-2\), and \(x=2\). It is further detailed that the function touches and turns around at \(x=1\), due to the even power of the corresponding factor, and crosses the x-axis at \(x=-2\) and \(x=2\).

Y-intercept of a function

Every polynomial function graph will cross the y-axis at a single point called the y-intercept. This intercept is found by evaluating the function at \(x=0\). The point where the graph crosses the y-axis provides a reference for graphing the entire function.

For our example, setting \(x=0\) in the function \(f(x)\) gives us \(f(0)=-3\), indicating that the y-intercept of the graph is at point \(0, -3\). This is a crucial checkpoint in graphing the polynomial, especially when plotting the initial shape of the curve.

Symmetry in graphs

Symmetry in graphs can tell us a great deal about the characteristics of polynomial functions. If a function's graph is symmetrical about the y-axis, it suggests even-function behavior, where \(f(x) = f(-x)\). If the graph is symmetrical with respect to the origin, it exhibits odd-function behavior, where \(f(x) = -f(-x)\).

For the graph of \(f(x)=-3(x-1)^2(x^2-4)\), testing for symmetry by replacing \(x\) with \(–x\) shows that it possesses neither y-axis symmetry nor origin symmetry. This finding enables us to understand that the shape of the graph will not mirror across any axis, which has to be accounted for when plotting points and creating the graph's visual representation.

Polynomial turning points

Turning points on a polynomial graph are where the function changes direction from increasing to decreasing (peaks) or from decreasing to increasing (troughs). According to the polynomial of degree \(n\), there can be at most \(n-1\) turning points.

In the function \(f(x)=-3(x-1)^{2}(x^{2}-4)\) of degree 4, we can anticipate up to 3 turning points. These points are crucial in plotting as they define the actual curvature and distinct shape of the graph. The step-by-step solution we have uses these potential turning points to verify the correctness of the graph, which in turn ensures a more accurate representation of the polynomial function.