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Chapter 2: Problem 59

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. $$f(x)=-x^{2}(x-1)(x+3)$$

Short Answer

Expert verified
The end behavior is 'fall-left, rise-right', x-intercepts are \(x = 0\), \(x = 1\), \(x = -3\), y-intercept is \(y = 0\). The graph has neither y-axis symmetry nor origin symmetry. Three of the turning points could potentially be at \(x = 0\), \(x = 1\), and \(x = -3\).

Step by step solution

01

Determine the End Behavior

Apply the Leading Coefficient Test. Here, the leading term is the highest degree term in the polynomial, \(x^{3}\) contributing to leading coefficient, \(-1\). Because the degree is odd and the leading coefficient is negative, the graph falls to the left and rises to the right.
02

Find the x-intercepts

Set the function equal to zero and solve for \(x\) to find the x-intercepts. -\(x^{2}(x-1)(x+3) = 0\). This gives the solutions \(x = 0\), \(x = 1\), and \(x = -3\). The graph crosses the x-axis at these points.
03

Find the y-intercept

Setting \(x = 0\) in the function yields the value of \(y\). So, \(f(0) = -0^{2}(0 - 1)(0+3) = 0\), which indicates that the y-intercept is \(0\).
04

Determine the Symmetry of the Graph

Replace \(x\) with \(-x\) and simplify. If we obtain the same function, then the graph has y-axis symmetry. If we obtain the negated function, then the graph has origin symmetry. If neither of these occur, the graph has no symmetry. For the function \(f(-x) = -(-x)^{2}(-x - 1)(-x + 3)\), we don't get the same function \(f(x)\) or \(-f(x)\). Therefore, the graph has neither y-axis symmetry nor origin symmetry.
05

Graph the Function and Check for Turning Points

Plot the x-intercepts and y-intercept, sketch the assumed graph of the function, correctly orienting it using the end behavior. For this cubic function, there could be up to 2 turning points. The x-intercepts at \(x = 0\), \(x = 1\), and \(x = -3\) can serve as turning points. We can confirm these with the graph. If necessary, perform additional checks to make certain of the location of the turning points.

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