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Chapter 2: Problem 47

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. $$f(x)=x^{4}-2 x^{3}+x^{2}$$

Short Answer

Expert verified
The graph of \(f(x)=x^{4}-2 x^{3}+x^{2}\) rises to the right and left. It touches the x-axis at \(x=0\) and crosses the x-axis at \(x=1\). The y-intercept is \(y=0\). The graph doesn't have either y-axis or origin symmetry. Plotting the graph confirms these properties.

Step by step solution

01

Determine the end behavior

The degree of the polynomial is 4, which is an even number and the leading coefficient is 1, which is positive. Hence by the Leading Coefficient Test, the graph rises to the right and rises to the left.
02

Find the x-intercepts

Set the polynomial equal to zero and solve for \(x\):\n\(x^{4}-2 x^{3}+x^{2} = 0\)\nThis simplifies to:\n\(x^{2}(x^{2}-2x+1) = 0\)\nIt can further be factored as: \n\(x^{2}(x-1)^{2} = 0\)\nThe roots or solutions are \(x=0, 1\). Because zero has a multiplicity of two, the graph touches the x-axis and turns around at x=0. For \(x=1\), the graph crosses the x-axis because the multiplicity is an even number, in this case, two.
03

Find the y-intercept

The y-intercept of the graph of the function is the value of the function at \(x=0\), which is \(f(0)=0\). So, the y-intercept is \(y=0\)
04

Check for symmetry

For symmetry about the y-axis, replace \(x\) by \(-x\), if we get the same function, it has y-axis symmetry. If not, then it doesn't. Replacing \(x\) by \(-x\) in \(f(x)=x^{4}-2 x^{3}+x^{2}\), the function does not remain the same, indicating that the graph is not symmetric about the y-axis.\n\nFor symmetry about the origin, replace \(x\) by \(-x\) and \(y\) by \(-y\), if we get the same function, it has origin symmetry. If not then it doesn't. Replacing \(y\) by \(-y\) and \(x\) by \(-x\) in \(f(x)=x^{4}-2 x^{3}+x^{2}\), the function does not remain the same, indicating that the graph is not symmetric about the origin.\n\nSo, the graph of the function has neither y-axis symmetry, nor origin symmetry.
05

Graph the function

Plot the points determined from the calculations for intercepts and test whether they completely describe the graph or whether we need extra points. For a degree 4 polynomial, we have at most 3 turning points, but since \(f(x)=x^{2}(x-1)^{2}\) has a repeated root, the graph is less complex. Splines between these points should suffice to accurately illustrate the graph.

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