Question

show me how to graph f(x)=x^3 - 2x^2

Explanation:

Step1: Find the \(x\) - intercepts

Set \(f(x)=0\), so \(x^{3}-2x^{2}=x^{2}(x - 2)=0\).
The solutions are \(x = 0\) (with multiplicity \(2\)) and \(x=2\).

Step2: Find the \(y\) - intercept

Set \(x = 0\), then \(f(0)=0^{3}-2\times0^{2}=0\).

Step3: Find the first derivative

Use the power rule. If \(y=x^{n}\), \(y^\prime=nx^{n - 1}\).
\(f^\prime(x)=3x^{2}-4x=x(3x - 4)\).
Set \(f^\prime(x)=0\), we get \(x = 0\) and \(x=\frac{4}{3}\).
When \(x<0\), \(f^\prime(x)>0\) (function is increasing).
When \(0When \(x>\frac{4}{3}\), \(f^\prime(x)>0\) (function is increasing).

Step4: Find the second derivative

\(f^{\prime\prime}(x)=6x-4\).
Set \(f^{\prime\prime}(x)=0\), then \(x=\frac{2}{3}\).
When \(x<\frac{2}{3}\), \(f^{\prime\prime}(x)<0\) (function is concave - down).
When \(x>\frac{2}{3}\), \(f^{\prime\prime}(x)>0\) (function is concave - up).

Step5: Plot key points and sketch the graph

Plot the \(x\) - intercepts \((0,0)\) and \((2,0)\), \(y\) - intercept \((0,0)\), critical points \((0,0)\) (local maximum since the function changes from increasing to decreasing at \(x = 0\)) and \((\frac{4}{3},f(\frac{4}{3}))\) where \(f(\frac{4}{3})=(\frac{4}{3})^{3}-2(\frac{4}{3})^{2}=\frac{64}{27}-\frac{32}{9}=\frac{64 - 96}{27}=-\frac{32}{27}\), and the inflection point \((\frac{2}{3},f(\frac{2}{3}))\) where \(f(\frac{2}{3})=(\frac{2}{3})^{3}-2(\frac{2}{3})^{2}=\frac{8}{27}-\frac{8}{9}=\frac{8 - 24}{27}=-\frac{16}{27}\). Then sketch the curve based on the increasing/decreasing and concavity information.

Answer:

Follow the above steps to graph the function \(y = x^{3}-2x^{2}\).