Problem 60 - 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. $$f(x)=-x^{2}(x+2)(x-2)$$

Step-by-Step Solution

Answer

The end behavior of the graph is: as \(x\) approaches positive infinity, \(f(x)\) approaches negative infinity; and as \(x\) approaches negative infinity, \(f(x)\) approaches negative infinity. The x-intercepts are -2, 0, and 2 and the function crosses the x-axis at each intercept. The y-intercept is at 0. The graph has no symmetry about the y-axis or the origin. The graph is a w-shaped curve that crosses the x-axis at \(-2\), \(0\), \(2\), and touches y-axis at (0,0).

Step 1: Apply the Leading Coefficient Test

Check the degree and the leading coefficient of the polynomial \(f(x)\). The degree is 4 since it's the sum of the exponent of \(x\) in each factor, and the leading coefficient is -1 because that's the coefficient on the \(x^{4}\) term. After determining these two factors, use the Leading Coefficient Test. Since here, the degree is even and the leading coefficient is negative, the end behavior of the graph will be: as \(x\) approaches positive infinity, \(f(x)\) approaches negative infinity; and as \(x\) approaches negative infinity, \(f(x)\) approaches negative infinity.

Step 2: Find the X-intercepts

To find the x-intercepts (also called roots or zeros), set \(f(x)\) equal to zero and solve for \(x\): \n0 = -x^{2}(x+2)(x-2)\nThis equation will be satisfied if any factor equals zero. So, the x-intercepts are \(-2\), \(0\), and \(2\). Because each factor \(x\), \(x+2\), and \(x-2\) has a multiplicity of 1 (meaning they appear only one time), the graph crosses the x-axis at each of these intercepts.

Step 3: Find the Y-intercept

To find the y-intercept, we have to substitute \(x=0\) into the function \(f(x)\): \nf(0) = -0^{2}(0+2)(0-2) = 0\nTherefore, the y-intercept is \(0\).

Step 4: Determine the symmetry

Check if the function is symmetric about the \(y\)-axis, the origin, or neither. To see if \(f(x)\) is symmetric about the y-axis, we substitute \(-x\) for \(x\) and simplify. If we get the same function that we started with, it is even (symmetry about the y-axis). If we get the negative of the function we started with, it is odd (symmetry about the origin). In this case \nf(-x) = -(-x)^{2}(-x+2)(-x-2)It is not equal to \(f(x)\), nor is it equal to \(-f(x)\), therefore, the graph has no symmetry.

Step 5: Graph the Function

Compute the function values at some x values within the range (-3,3) and plot those points. Pay special attention to the values of \(x\) close to each intercept. The graph will be a w-shaped curve that crosses the x-axis at -2, 0, 2 and touches y-axis at (0,0). It trends down as \(x\) approaches positive and negative infinity.

Key Concepts

Leading Coefficient Test

The Leading Coefficient Test helps us understand the end behavior of polynomial graphs by examining the highest degree term. For the function \(f(x) = -x^2(x+2)(x-2)\), the highest degree term is found by multiplying the largest power of \(x\) in each factor. Here, \(-x^2(x^1)(x^1) = -x^4\), so the degree is 4. The leading coefficient is \(-1\) from \(-x^4\).

The leading coefficient tells us the direction the graph will head as \(x\) approaches infinity. Since the degree is even and the leading coefficient is negative, the graph falls to negative infinity on both the far left and far right ends. In simpler terms, it swoops down both on the left and right sides.

Intercepts

Intercepts are key points where the graph meets the x-axis and y-axis. The x-intercepts, or roots, occur when the graph touches or crosses the x-axis. To find them, set \(f(x) = 0\):
- Factoring out, we have \(-x^2(x+2)(x-2) = 0\). The solutions are obtained by setting each factor equal to zero:
- \(x = 0\)
- \(x+2 = 0\), thus \(x = -2\)
- \(x-2 = 0\), thus \(x = 2\)
These are the x-intercepts and the graph crosses at these points because each has a multiplicity of 1.

Finding the y-intercept is straightforward: substitute \(x = 0\) into the original function. Here, \(f(0) = -0^2(0+2)(0-2) = 0\), so the y-intercept is at \(0\).

Graph Symmetry

Determining symmetry simplifies graphing. Symmetry about the y-axis happens if replacing \(x\) with \(-x\) gives the original function: \(f(-x) = f(x)\). Origin symmetry occurs if \(f(-x) = -f(x)\).

For \(f(x) = -x^2(x+2)(x-2)\), check symmetry by substituting \(-x\) and comparing:
- \(f(-x) = -(-x)^2(-x+2)(-x-2)\). Simplifying doesn't result in \(f(x)\) or \(-f(x)\), confirming no symmetry exists about the y-axis or the origin. This means, in terms of symmetry, the graph is neither even nor odd.

Degree of a Polynomial

The degree of a polynomial is the sum of the highest powers of all variables in its terms. It's crucial in determining the graph's shape and end behavior. For \(f(x) = -x^2(x+2)(x-2)\), we compute: sum the exponents to find the degree of large term of each factor.
- Multiplying leads to \(-x^4\), yielding a degree of 4.

Degrees dictate maximum possible turning points and basic graph shape. With degree 4, expect UP TO 3 turning points. However, note the real turning points can be fewer, but never more. For any polynomial, the degree also equals the number of x-intercepts (roots) once counted considering their multiplicity. Here, there are 3 real roots, matching expectations for such a polynomial graph.