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Sketching the Graph of a Polynomial Function Sketch the graph of the function by (a) applying the Leading Coefficient Test, (b) finding the real zeros of the polynomial, (c) plotting sufficient solution points, and (d) drawing a continuous curve through the points. $$h(x)=\frac{1}{3} x^{3}(x-4)^{2}$$

Short Answer

Expert verified
The graph of the function \(h(x) = \frac{1}{3} x^{3}(x-4)^{2}\) starts from the bottom, sharply rises to intersect the x-axis at \(x=0\), falls below the x-axis again, touches the x-axis at \(x=4\), before rising sharply again. Behaviour at extreme ends is towards \(\pm \infty\).

Step by step solution

01

Applying the Leading Coefficient Test

The function \(h(x) = \frac{1}{3} x^{3}(x-4)^{2}\) has degree 5 and a positive leading coefficient, \(\frac{1}{3}\). Therefore, according to the leading coefficient test, the end behaviour of the function will be that as \(x \rightarrow -\infty\), \(h(x) \rightarrow -\infty\) and as \(x \rightarrow \infty\), \(h(x) \rightarrow \infty\).
02

Finding the Real Zeros of the Polynomial

To find the real zeros of the polynomial, we must figure out what values of \(x\) cause \(h(x)\) to equal zero. That means, we need to solve the equation \(\frac{1}{3} x^{3}(x-4)^{2} = 0\). This equation gives us two real zeros, \(x=0\) and \(x=4\). However, since \(x=4\) originated as a square, it has multiplicity 2, hence the graph will intersect the x-axis at \(x=0\) and merely touch the x-axis at \(x=4\).
03

Plotting Sufficient Solution Points

Now we need to select some points and compute their corresponding \(h(x)\) values. This action will assist us in drawing the graph. Choose points that are in the regions split by the x-intercepts (like -1, 2, and 5). After calculating, we get \(-1 \rightarrow -\frac{15}{3}\), \(2 \rightarrow -\frac{16}{3}\) and, \(5 \rightarrow \frac{45}{3}\).
04

Drawing the Continuous Curve

Lastly, combine the information from the previous steps to sketch the graph. Draw a smooth, continuous curve through all points, making sure to intersect the x-axis at \(x=0\), touch the x-axis at the double zero \(x=4\), and follow the end behaviour determined in Step 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Leading Coefficient Test
Understanding how the leading coefficient influences the graph of a polynomial function is crucial for sketching its behavior. The Leading Coefficient Test allows us to predict the 'end behavior' of a polynomial graph merely based on its degree and the sign of its leading coefficient.

For even-degree polynomials with a positive leading coefficient, the graph will rise on both ends. Conversely, if the leading coefficient is negative, the graph will fall on both ends. However, in the case of odd-degree polynomials, if the leading coefficient is positive, the graph will fall as the input approaches negative infinity and rise as the input approaches positive infinity. For a negative leading coefficient and an odd degree, the graph will do the opposite.

Take, for example, the function given in the exercise, \(h(x) = \frac{1}{3} x^{3}(x-4)^{2}\). It is a polynomial of degree 5 (odd), with a positive leading coefficient \(\frac{1}{3}\). According to the test, the graph falls to the left (as \(x \rightarrow -\infty\), \(h(x) \rightarrow -\infty\)) and rises to the right (as \(x \rightarrow \infty\), \(h(x) \rightarrow \infty\)). This initial insight guides us on how to start and finish drawing the function's graph.
Real Zeros of Polynomial
The real zeros of a polynomial are the x-values where the polynomial evaluates to zero. These are the points where the graph intersects the x-axis. Finding real zeros is not just about locating where the graph hits the axis; it also informs us about the shape of the graph at those intersections.

In the context of the provided exercise function \(h(x) = \frac{1}{3} x^{3}(x-4)^{2}\), solving \(\frac{1}{3} x^{3}(x-4)^{2} = 0\) reveals zeros at \(x=0\) and \(x=4\). However, the zero at \(x=4\) is a double zero due to its square (has a multiplicity of 2), which means the graph only touches the x-axis there without crossing it. In contrast, at \(x=0\) the graph does intersect the x-axis. Multiplicity affects the graph's behavior: zeros with even multiplicity result in the graph touching the x-axis, while zeros with odd multiplicity ensure crossing.
Graphing Polynomial Functions
When it comes to graphing polynomial functions, we combine our knowledge from the Leading Coefficient Test and the location of real zeros to create an accurate sketch. To refine our graph, we calculate the values of the function at various x-values, especially around the zeros and transition areas of the polynomial.

For the given function, after determining the end behavior and zeros, we select additional points such as -1, 2, and 5. Calculating the y-values for these points provides us with crucial 'checkpoints', guiding how the curve moves between the zeros and in accordance to the end behavior.

Graphing involves plotting the calculated points and zeros, then drawing smooth curves that connect these dots while respecting the behavior dictated by the Leading Coefficient Test. The goal is to produce a continuous curve that accurately reflects the polynomial's characteristics, ensuring it passes through or touches the x-intercepts and aligns with the predicted rise and fall at the graph’s ends.

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Most popular questions from this chapter

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