Problem 60 - College Algebra

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 uning points to check whether it is drawn correctly. $$ f(x)=-x^{2}(x+2)(x-2) $$

Step-by-Step Solution

Answer

End behavior: As \(x\) approaches \(\pm\infty\), \(f(x)\) approaches \(-\infty\). \(X\)-intercepts: 0, 2, -2. \(Y\)-intercept: 0. The graph has \(y\)-axis symmetry. Refer to graph for step 5.

Step 1: End Behavior using Leading Coefficient Test

The degree of the polynomial is 4 and the leading coefficient is -1 (from \(-x^{2}\)). If the degree is even, and the leading coefficient is negative, as per the Leading Coefficient Test, the end behavior of the polynomial is: As \(x\) approaches positive infinity (\(+\infty\)), \(f(x)\) approaches positive infinity (\(+\infty\)), and as \(x\) approaches negative infinity (\(-\infty\)), \(f(x)\) approaches positive infinity (\(+\infty\)).

Step 2: Finding \(x\)-Intercepts

To find the \(x\)-intercepts, set \(f(x)\) equal to zero and solve for \(x\). \(-x^{2}(x+2)(x-2) = 0\). This gives \(x\) = 0, \(x\) = 2, and \(x\) = -2. The graph crosses the \(x\)-axis at these intercepts, since the multiplicity of the roots is 1 (odd).

Step 3: Finding the \(y\)-Intercept

To find the \(y\)-intercept, set \(x\) to 0 in the function, which gives \(f(0) = -0^2*(0+2)*(0-2) = 0\). Therefore, the \(y\)-intercept is \(0\).

Step 4: Determine Symmetry

The given function is even (=4), hence the graph will have \(y\)-axis symmetry.

Step 5: Graphing the function

Plot the points obtained from steps 2 and 3 and sketch based on these points and the end behavior. The turning points of the graph should be less than or equal to 4-1 = 3, as per theorem of polynomial function.

Key Concepts

Leading Coefficient Test

The Leading Coefficient Test helps predict the end behavior of polynomial functions by examining the degree of the polynomial and its leading coefficient.
The degree of a polynomial is the highest power of the variable in the equation. The leading coefficient is the coefficient of this term.
For example, in the polynomial function \( f(x) = -x^{2}(x+2)(x-2) \), the highest power of \( x \) is 4, making it a fourth-degree polynomial, and the leading coefficient is \(-1\).
Given these factors:

• If the degree is even, such as 4, and the leading coefficient is negative (\(-1\)), both ends of the graph head upwards as \( x \) approaches both infinity. Essentially, the graph will resemble a downward-opening U shape.
• If the degree was odd, or the leading coefficient positive, the end behavior would differ, showcasing diverging trends at each end of the graph.

x-Intercepts

x-intercepts are points where the graph of a function crosses or touches the x-axis. At these points, the function's value is zero.
To find the x-intercepts for our polynomial \( f(x) = -x^{2}(x+2)(x-2) \), set the function equal to zero and solve:

• \(-x^{2}(x+2)(x-2)=0_{ \)\( x=0, x=2, x=-2 \)}.

The graph crosses the x-axis at these points. The reason it crosses, and not just touches, is due to the odd multiplicity of the roots (which is 1 in this case).
If any root had an even multiplicity, the graph would simply touch the x-axis and turn around at that intercept.

y-Intercept

Finding the y-intercept involves setting \( x \) to zero and solving for the function value.
For the equation \( f(x) = -x^{2}(x+2)(x-2) \), substitute \( x = 0 \):

• \( f(0) = -0^{2}(0+2)(0-2) = 0 \).

Thus, the y-intercept of the function is at the origin, (0, 0).
The y-intercept represents the point where the graph crosses the y-axis, indicating the value of \( f(x) \) when all terms containing \( x \) are zero.

Graph Symmetry

Graph symmetry concerns how a graph is mirrored across certain lines or points.
In this polynomial function, determining symmetry is tied to its degree and form.
The polynomial \( f(x) = -x^{2}(x+2)(x-2) \) is an even-degree polynomial (degree of 4).
This tells us the graph should have y-axis symmetry:

• Y-axis symmetry implies folding the graph along the y-axis mirrors it perfectly, meaning the graph is symmetric about the vertical axis at \( x = 0 \).

If the polynomial were of an odd degree, we might expect origin symmetry instead.
Origin symmetry involves a 180-degree rotation around the origin making the graph appear unchanged.
It's important to note whether symmetry exists before sketching as it helps predict graph behavior.