Problem 49 - College Algebra

Question

In Exercises \(35-50\) a. Use the Leading Coefficient Test to determine the graphs end behavior. b. Find \(x\) -intercepts by setting \(f(x)=0\) and solving the resulting polynomial equation. State whether the graph crosses the \(x\)-axis, or touches the \(x\)-axis and turns around, at each intercept. c. Find the \(y\) -intercept by setting \(x\) equal to 0 and computing \(f(0)\) 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 fact that the maximum number of turning points of the graph is \(n-1\) 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 function is that the function falls to both sides. The graph of the function crosses the x-axis at \(x=1\), \(x=-2\), and \(x=2\). It has its y-intercept at -12. The graph is neither symmetric about the y-axis nor the origin. The function graphed correctly should have maximum of 3 turning points.

Step 1: Determine the end behavior

By analyzing the leading term (the term with the highest degree), which is \(-3x^4\), one can conclude, using the Leading Coefficient Test, that as \(x\) approaches negative or positive infinity, \(f(x)\) will also go towards negative infinity. Therefore, the graph falls to the right and to the left.

Step 2: Find the x-intercepts

To find the \(x\) intercepts, set the function equal to zero and solve for \(x\). \(-3(x-1)^{2}(x^{2}-4) = 0\) give the solutions \(x=1, -2, 2\). The graph crosses the \(x\)-axis at \(x=1\), \(x=2\) and \(x=-2\).

Step 3: Find the y-intercept

To find the \(y\) intercept, set \(x\) equal to zero in the original function. This gives \(f(0) = -3(0-1)^{2}(0^{2}-4) = -12\). Therefore, the \(y\)-intercept of the graph is -12.

Step 4: Determine the symmetry

Replace \(x\) with \(-x\) in the function and simplify. If \(-f(x)\) is achieved then the graph is symmetric about the y-axis, if \(f(x)\) is achieved then the graph is symmetric about the origin. Neither of these are true for the given function, hence it has neither y-axis symmetry, nor origin symmetry.

Step 5: Graph the function

Plot the points found above and draw a smooth curve through these points, making sure that the end behavior and symmetry characteristics are correct. The maximum number of turning points of a polynomial function is \(n-1\), where \(n\) is the degree of the polynomial. Here the degree is 4, so maximum number of turning points is \(4-1=3\).

Key Concepts

Leading Coefficient Test

The Leading Coefficient Test helps us predict how a polynomial graph will behave at the far ends of the x-axis, known as its "end behavior." For the function \(f(x)=-3(x-1)^{2}(x^{2}-4)\), we focus on the term with the highest degree, which is \(-3x^4\). Here, the degree of the polynomial is 4, and the leading coefficient is \(-3\).

This tells us:

• Since the degree is even (4), the end behavior will be similar on both sides of the graph.
• The leading coefficient is negative, so as \(x\) approaches both negative and positive infinity, \(f(x)\) will head towards negative infinity. This means the graph "falls" on both ends.

Understanding this helps in sketching the overall shape, showing both ends going downward.

End Behavior

End behavior refers to how a graph behaves as \(x\) approaches infinity or negative infinity. For our polynomial, with a degree of 4 and a negative leading coefficient, the end behavior is straightforward.

• As \(x\) approaches infinity \((x \to +\infty)\), \(f(x) \to -\infty\).
• As \(x\) approaches negative infinity \((x \to -\infty)\), \(f(x) \to -\infty\).

This means the graph will descend on both the left and right sides. Recognizing end behavior is crucial for sketching a graph, giving it context in the larger picture.

X-intercepts

X-intercepts occur where the graph crosses the x-axis, meaning \(f(x) = 0\). For \(-3(x-1)^{2}(x^{2}-4) = 0\), we solve the equation to find the intercepts:

• \(x - 1 = 0 \Rightarrow x = 1\)
• \(x^2 - 4 = 0 \Rightarrow x = -2, 2\)

The graph crosses the x-axis at \(x = 1\), \(x = -2\), and \(x = 2\). At each intercept:

• The term \((x-1)^{2}\) means the graph "touches" the x-axis at \(x = 1\) without crossing, creating a "bounce."
  
• At \(x = -2\) and \(x = 2\), the graph actually crosses the axis.

X-intercepts help locate where the graph interacts with the x-axis, a key characteristic for plotting.

Y-intercept

A y-intercept is where the graph crosses the y-axis. This happens when you set \(x = 0\) in the equation. For \(-3(x-1)^{2}(x^{2}-4)\), substituting \(x = 0\) gives:

\[f(0) = -3(0-1)^{2}(0^{2}-4) = -12\]
Thus, the y-intercept is \(-12\). This is the single point where the graph crosses the y-axis, providing an anchor point for drawing the graph.
The y-intercept offers a starting reference point, making it an essential element when sketching or understanding the movement of the graph.