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Chapter 2: Problem 88

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. $$g(x)=\frac{1}{10}(x+1)^{2}(x-3)^{3}$$

Short Answer

Expert verified
The function graph will start downward on the left, touch the x-axis at x=-1, then pass through x=3 and go upward. The points plotted will include the zeros x=-1, x=3 and other points around these zeros that we calculated by substituting those x-values in the function.

Step by step solution

01

Applying the Leading Coefficient Test

The leading coefficient of the function \(g(x) = \frac{1}{10}(x+1)^{2}(x-3)^{3}\) is \(+\frac{1}{10}\). And the degree of the function is \(2+3 = 5\), which is odd. According to the Leading Coefficient Test, if the leading coefficient is positive and degree is odd, the end behavior of the graph is: as \(x \rightarrow -\infty, g(x) \rightarrow -\infty\) and as \(x \rightarrow +\infty, g(x) \rightarrow +\infty\). That means, the tail of the graph on the left tends downwards, while on the right, it tends upwards.
02

Finding the Real Zeros of the Polynomial

The real zeros of the polynomial can be calculated by equating the function to zero and solving for \(x\). For \(g(x) = \frac{1}{10}(x+1)^{2}(x-3)^{3} = 0\), the real zeros are \(x = -1, 3\). Given the multiplicity of the roots, \(x = -1\) is of multiplicity 2 and \( x = 3\) is of multiplicity 3, the curve will touch and turn around at \(x = -1\) and pass through \(x = 3\).
03

Plotting Sufficient Solution Points

Now, we will select additional points around the zeros for plotting. Suggested points are \(x = -2, 0, 1, 2, 4\). We can substitute these values in the function to get their respective \(y\) values. We plot these points along with the zeros on the graph.
04

Drawing a Continuous Curve through the Points

Finally, we draw a smooth curve that goes through all the plotted points and zeros. Ensure the curve goes upwards at \(x = 3\) and downwards at \(x = -1\) as per the Leading Coefficient Test and the behavior at the zeros, while making sure the curve is continuous and smooth.

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Most popular questions from this chapter

Use Descartes's Rule of Signs to determine the possible numbers of positive and negative real zeros of the function. $$h(x)=2 x^{3}+3 x^{2}+1$$

The maximum safe load uniformly distributed over a one-foot section of a two- inch-wide wooden beam can be approximated by the model $$\text { Load }=168.5 d^{2}-472.1$$ where \(d\) is the depth of the beam. (a) Evaluate the model for \(d=4, d=6, d=8, d=10\) and \(d=12 .\) Use the results to create a bar graph. (b) Determine the minimum depth of the beam that will safely support a load of 2000 pounds.

Explore transformations of the form \(g(x)=a(x-h)^{5}+k\) (a) Use a graphing utility to graph the functions \(y_{1}=-\frac{1}{3}(x-2)^{5}+1\) and \(y_{2}=\frac{3}{5}(x+2)^{5}-3\) Determine whether the graphs are increasing or decreasing. Explain. (b) Will the graph of \(g\) always be increasing or decreasing? If so, then is this behavior determined by \(a, h,\) or \(k ?\) Explain. (c) Use the graphing utility to graph the function \(H(x)=x^{5}-3 x^{3}+2 x+1\) Use the graph and the result of part (b) to determine whether \(H\) can be written in the form \(H(x)=a(x-h)^{5}+k\) Explain.

Match the cubic function with the numbers of rational and irrational zeros. (a) Rational zeros: \(0 ;\) irrational zeros: 1 (b) Rational zeros: \(3 ;\) irrational zeros: 0 (c) Rational zeros: \(1 ;\) irrational zeros: 2 (d) Rational zeros: \(1 ;\) irrational zeros: 0 $$f(x)=x^{3}-2 x$$

Find all the zeros of the function. When there is an extended list of possible rational zeros, use a graphing utility to graph the function in order to disregard any of the possible rational zeros that are obviously not zeros of the function. $$f(x)=9 x^{3}-15 x^{2}+11 x-5$$
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