Question: User: f(x)=x(x-2)(x+1)^2 answer the questions regarding the graph of f. Then, use this information to graph the function. (a) Choose the end behavior of the graph of f. (b) List each real zero of f according to the behavior of the graph at the x-axis near that zero. If there is more than one answer, separate them with commas. If there is no answer, click on "None". Zero(s) where the graph crosses the x-axis: Zero(s) where the graph touches, but does not cross the x-axis: (c) Find the y-intercept of the graph of f. y-intercept: (d) Graph f(x)=x(x-2)(x+1)^2 by doing the following. Polynomial and Rational Functions Determining end behavior and intercepts to graph a polynomial function Click on the graph-a-function button. Graph shape/math position_hint: top-left; polynomial f(x)=x(x-2)(x+1)^2 with zeros at x=-1 (double root, touches/bounces), x=0 and x=2 (simple roots, crosses), y-intercept at (0,0), and end behavior down on the left and up on the right.

1. **State the problem:** We analyze the polynomial function $$f(x)=x(x-2)(x+1)^2$$ to determine its end behavior, zeros and their behavior at the x-axis, y-intercept, and then graph it. 2. **End behavior:** The degree of $$f(x)$$ is $$1 + 1 + 2 = 4$$ (even degree) and the leading coefficient is positive (from $$x \cdot x \cdot x^2 = x^4$$ with positive coefficient). For even degree polynomials with positive leading coefficient, the end behavior is: $$\text{As } x \to -\infty, f(x) \to +\infty$$ $$\text{As } x \to +\infty, f(x) \to +\infty$$ However, let's confirm by expanding the leading term: $$f(x) = x(x-2)(x+1)^2 = x(x-2)(x^2 + 2x + 1)$$ Leading term: $$x \cdot x \cdot x^2 = x^4$$ Positive leading coefficient means both ends go up. 3. **Zeros and their behavior:** - Zeros are at $$x=0$$, $$x=2$$, and $$x=-1$$. - Multiplicity: - $$x=0$$ has multiplicity 1 (linear factor $$x$$), so graph crosses the x-axis here. - $$x=2$$ has multiplicity 1 (linear factor $$x-2$$), so graph crosses the x-axis here. - $$x=-1$$ has multiplicity 2 (factor $$(x+1)^2$$), so graph touches and bounces off the x-axis here. 4. **Y-intercept:** Evaluate $$f(0)$$: $$f(0) = 0 \cdot (0-2) \cdot (0+1)^2 = 0$$ So the y-intercept is at $$(0,0)$$. 5. **Summary:** - End behavior: $$f(x) \to +\infty$$ as $$x \to \pm \infty$$. - Zeros crossing x-axis: $$x=0, 2$$. - Zero touching x-axis: $$x=-1$$. - Y-intercept: $$(0,0)$$. 6. **Graphing:** - At $$x=-1$$, the graph touches and bounces off the x-axis. - At $$x=0$$ and $$x=2$$, the graph crosses the x-axis. - The graph goes up on both ends. This matches the description: zeros at $$x=-1$$ (double root, touches/bounces), $$x=0$$ and $$x=2$$ (simple roots, crosses), y-intercept at $$(0,0)$$, and end behavior up on both sides. **Final answers:** (a) End behavior: $$f(x) \to +\infty$$ as $$x \to -\infty$$ and $$f(x) \to +\infty$$ as $$x \to +\infty$$. (b) Zero(s) where the graph crosses the x-axis: $$0, 2$$. Zero(s) where the graph touches, but does not cross the x-axis: $$-1$$. (c) Y-intercept: $$(0,0)$$. (d) Graph as described above.