Question

show me how to graph f (x) = x3 - 4x2

Explanation:

Step1: Find the \(x\) - intercepts

Set \(y = f(x)=x^{3}-4x^{2}=x^{2}(x - 4)\).
When \(y = 0\), \(x^{2}(x - 4)=0\).
Using the zero - product property \(a\times b = 0\Rightarrow a = 0\) or \(b = 0\).
We get \(x=0\) (a double root) and \(x = 4\). So the \(x\) - intercepts are \((0,0)\) and \((4,0)\).

Step2: Find the \(y\) - intercept

Set \(x = 0\). Then \(y=f(0)=0^{3}-4\times0^{2}=0\). So the \(y\) - intercept is \((0,0)\).

Step3: Find the first derivative

Use the power rule \((x^{n})^\prime=nx^{n - 1}\).
\(f^\prime(x)=3x^{2}-8x=x(3x - 8)\).
Set \(f^\prime(x)=0\), then \(x(3x - 8)=0\).
Solving \(x(3x - 8)=0\) gives \(x = 0\) and \(x=\frac{8}{3}\).
When \(x\lt0\), let \(x=-1\), \(f^\prime(-1)=3\times(-1)^{2}-8\times(-1)=3 + 8=11\gt0\), so \(f(x)\) is increasing on \((-\infty,0)\).
When \(0\lt x\lt\frac{8}{3}\), let \(x = 1\), \(f^\prime(1)=3\times1^{2}-8\times1=3 - 8=-5\lt0\), so \(f(x)\) is decreasing on \((0,\frac{8}{3})\).
When \(x\gt\frac{8}{3}\), let \(x = 3\), \(f^\prime(3)=3\times3^{2}-8\times3=27-24 = 3\gt0\), so \(f(x)\) is increasing on \((\frac{8}{3},\infty)\).
The local maximum is \(f(0)=0\) and the local minimum is \(f(\frac{8}{3})=(\frac{8}{3})^{3}-4\times(\frac{8}{3})^{2}=\frac{512}{27}-\frac{256}{9}=\frac{512 - 768}{27}=-\frac{256}{27}\approx - 9.48\).

Step4: Find the second derivative

\(f^{\prime\prime}(x)=6x-8\).
Set \(f^{\prime\prime}(x)=0\), then \(6x-8 = 0\), \(x=\frac{4}{3}\).
When \(x\lt\frac{4}{3}\), let \(x = 0\), \(f^{\prime\prime}(0)=6\times0-8=-8\lt0\), so the graph is concave down on \((-\infty,\frac{4}{3})\).
When \(x\gt\frac{4}{3}\), let \(x = 2\), \(f^{\prime\prime}(2)=6\times2-8 = 4\gt0\), so the graph is concave up on \((\frac{4}{3},\infty)\). The inflection point is \(f(\frac{4}{3})=(\frac{4}{3})^{3}-4\times(\frac{4}{3})^{2}=\frac{64}{27}-\frac{64}{9}=\frac{64 - 192}{27}=-\frac{128}{27}\approx - 4.74\).

Step5: Plot additional points

Choose some \(x\) values, for example:
When \(x = 1\), \(y=f(1)=1^{3}-4\times1^{2}=1 - 4=-3\).
When \(x = 2\), \(y=f(2)=2^{3}-4\times2^{2}=8 - 16=-8\).
When \(x=5\), \(y=f(5)=5^{3}-4\times5^{2}=125-100 = 25\).

Answer:

To graph \(y = x^{3}-4x^{2}\):

• Mark the intercepts \((0,0)\) and \((4,0)\).
• Use the first - derivative information to identify where the function is increasing (\((-\infty,0)\cup(\frac{8}{3},\infty)\)) and decreasing (\((0,\frac{8}{3})\)), and local extrema \((0,0)\) (local maximum) and \((\frac{8}{3},-\frac{256}{27})\) (local minimum).
• Use the second - derivative information to identify concavity (concave down on \((-\infty,\frac{4}{3})\), concave up on \((\frac{4}{3},\infty)\)) and the inflection point \((\frac{4}{3},-\frac{128}{27})\).
• Plot additional points like \((1,-3)\), \((2,-8)\), \((5,25)\) and connect the points with a smooth curve.