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

Explanation:

Step1: Find the x - intercepts

Factor the function (f(x)=-x^{3}+4x^{2}-3x=-x(x^{2}-4x + 3)=-x(x - 1)(x - 3)). Set (y = 0), then (-x(x - 1)(x - 3)=0). Using the zero - product property (a\times b\times c = 0) implies (a = 0) or (b = 0) or (c = 0). So (x=0), (x = 1), (x = 3) are the x - intercepts.

Step2: Find the y - intercept

Set (x = 0) in (y=-x^{3}+4x^{2}-3x). Then (y=f(0)=0).

Step3: Analyze the end - behavior

Since the degree of the polynomial (y=-x^{3}+4x^{2}-3x) is (n = 3) (odd) and the leading coefficient (a=-1\lt0). As (x\rightarrow-\infty), (y=-x^{3}+4x^{2}-3x\approx -x^{3}\rightarrow+\infty) (because (-(-\infty)^{3}=+\infty)). As (x\rightarrow+\infty), (y=-x^{3}+4x^{2}-3x\approx -x^{3}\rightarrow-\infty).

Step4: Find the critical points

Differentiate (y=-x^{3}+4x^{2}-3x) using the power rule ((x^{n})^\prime=nx^{n - 1}). (y^\prime=-3x^{2}+8x - 3). Set (y^\prime = 0), then (3x^{2}-8x + 3=0). Using the quadratic formula (x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) for (ax^{2}+bx + c = 0) (here (a = 3), (b=-8), (c = 3)). (x=\frac{8\pm\sqrt{64-36}}{6}=\frac{8\pm\sqrt{28}}{6}=\frac{8\pm2\sqrt{7}}{6}=\frac{4\pm\sqrt{7}}{3}\approx\frac{4\pm2.65}{3}). (x_1=\frac{4+\sqrt{7}}{3}\approx2.22), (x_2=\frac{4-\sqrt{7}}{3}\approx0.78).

Step5: Analyze the sign of the derivative

Take test intervals: For (x\lt\frac{4 - \sqrt{7}}{3}\approx0.78), let (x = 0), then (y^\prime(0)=-3(0)^{2}+8(0)-3=-3\lt0). For (\frac{4 - \sqrt{7}}{3}\lt x\lt\frac{4+\sqrt{7}}{3}\approx2.22), let (x = 1), then (y^\prime(1)=-3(1)^{2}+8(1)-3=2\gt0). For (x\gt\frac{4+\sqrt{7}}{3}), let (x = 3), then (y^\prime(3)=-3(3)^{2}+8(3)-3=-6\lt0).

Answer:

Plot the x - intercepts ((0,0)), ((1,0)), ((3,0)), y - intercept ((0,0)). As (x\rightarrow-\infty,y\rightarrow+\infty) and as (x\rightarrow+\infty,y\rightarrow-\infty). The function is decreasing on ((-\infty,\frac{4 - \sqrt{7}}{3})) and ((\frac{4+\sqrt{7}}{3},+\infty)), increasing on ((\frac{4 - \sqrt{7}}{3},\frac{4+\sqrt{7}}{3})). Sketch the curve passing through these points and with the determined behavior.