Problem 60 - Algebra and Trigonometry

Question

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

End behavior: as \( x \) goes to \( +\infty \) and \( -\infty \), \( f(x) \) goes to \( -\infty \). X-intercepts: \( x = -2, 0, 2 \) and the graph crosses the \( x \)-axis at each intercept. Y-intercept is 0. The graph has origin symmetry and no y-axis symmetry.

Step 1: Determine End Behavior Using the Leading Coefficient Test

The degree of the polynomial \( f(x) = -x^{2}(x+2)(x-2) \) is 4 and the leading coefficient is negative. According to the Leading Coefficient Test, even degree polynomials with a positive leading coefficient rise to the left and right; when the leading coefficient is negative, the graph falls to the left and right. So, the end behavior of this function is: As \( x \) approaches \( +\infty \), \( f(x) \) approaches \( -\infty \) and as \( x \) approaches \( -\infty \), \( f(x) \) approaches \( -\infty \).

Step 2: Find the x-intercepts

The x-intercepts are the values of \( x \) that make the function \( f(x) \) equal to zero: \( f(x) = -x^{2}(x+2)(x-2) = 0 \). Solving this equation gives \( x = -2, 0, 2 \). At each of these points, the graph crosses the \( x \)-axis because the multiplicities of these zeroes are odd (1, 2, and 1 respectively).

Step 3: Find the y-intercept

The y-intercept is the value of the function at \( x = 0 \), that is, \( f(0) = -0^{2}(0+2)(0-2) = 0 \), so the y-intercept is 0.

Step 4: Determine Symmetry

Substitute \( -x \) for \( x \) in the function: \( f(-x) = -(-x)^{2}(-x+2)(-x-2) = -x^{2}(x-2)(x+2) \). Because \( f(-x) \) is not equivalent to \( f(x) \), there is no y-axis symmetry. Since \( f(-x) \) is equivalent to \( -f(x) \), there is origin symmetry.

Step 5: Graph the Function

To access the completeness of the graph, use the maximum number of turning points of the function, which is equal to the degree of the function minus one. Here, that equals to 4-1=3. Mark the \( x \)-intercepts at \( x = -2, 0, 2 \) and the \( y \)-intercept at \( y = 0 \). Draw the graph starting from 'falling' at \( -\infty \), crossing the \( x \)-axis at \( x = -2, 0, 2 \), still 'falling' to \( +\infty \), and displaying 3 turning points.

Key Concepts

Leading Coefficient Test

Understanding the Leading Coefficient Test can give you a quick snapshot of a polynomial function's end behavior—how the function behaves as the input values become very large or very small. This test revolves around two main components: the degree of the polynomial and the leading coefficient.

The coefficient of the term with the highest power in the polynomial is the 'leading' coefficient. For example, in the function f(x) = -x^4 + 5x^3 - 2x + 7, the leading coefficient is '-1', associated with the x^4 term.

When we apply this test:

• An even degree polynomial with a positive leading coefficient will rise on both ends.
• An even degree polynomial with a negative leading coefficient will fall on both ends.
• An odd degree polynomial with a positive leading coefficient will fall on the left and rise on the right.
• An odd degree polynomial with a negative leading coefficient will rise on the left and fall on the right.

So, with a function like f(x)=-x^2(x+2)(x-2), which has an even degree and a negative leading coefficient, we anticipate that the graph will point downwards as x approaches both +fty and -fty.

x-intercepts of Polynomial

The x-intercepts are the points where the polynomial function intersects the x-axis. For the function f(x), these are the values of x that make f(x) = 0. To find them, you'll want to set the entire polynomial equal to zero and solve for the roots.

In our function f(x)=-x^2(x+2)(x-2), we find three x-intercepts at x = -2, 0, and 2. Each intercept carries with it a behavior—whether the graph 'crosses' or 'touches' the x-axis at these points. This is where multiplicity plays a role:

• If a root is repeated (an even multiplicity), the graph will touch the x-axis and turn around.
• If a root is not repeated (an odd multiplicity), the graph will cross the x-axis.

Since our intersections have multiplicities of 1 and 2, we see the graph crossing the x-axis at x = -2 and x = 2, and touching at x = 0 as the multiplicity there is 2.

y-intercept of Function

The y-intercept is the point where the graph crosses the y-axis. To find it, simply substitute x=0 into the function. The result is the output f(0), which gives the y-coordinate of the y-intercept.

In our equation f(x)=-x^2(x+2)(x-2), when x is zero, all terms containing x vanish and thus f(0) = 0. This tells us that the y-intercept is 0, or in other words, the graph goes through the origin (0, 0). The y-intercept is a useful starting point when sketching a graph by hand.

Symmetry in Graphs

Graphs of polynomial functions can exhibit symmetry, which simplifies their analysis and graphing. We typically look for two types of symmetry: y-axis symmetry and origin symmetry. To check for y-axis symmetry, we replace x with -x and see if the resulting function f(-x) matches the original f(x). If it does, the graph is symmetric about the y-axis.

To check for symmetry about the origin, if f(-x) equals -f(x), as is the case in our function f(x)=-x^2(x+2)(x-2), then the graph has origin symmetry—it behaves the same way in all four quadrants relative to the origin.

Knowing whether a graph has symmetry can be very helpful. It means that you only need to calculate half of the graph, and then you can reflect those points across the axis of symmetry or rotate them about the origin to fill out the rest of the graph.