Question

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

Explanation:

Step1: Find the x - intercepts

Set \(y = f(x)=0\).
By the zero - product property, if \((x + 1)^{2}(x - 1)(x - 3)=0\), then \(x=-1\) (with multiplicity \(m = 2\)), \(x = 1\) (with multiplicity \(m=1\)), and \(x = 3\) (with multiplicity \(m = 1\)).

Step2: Determine the end - behavior

The leading term of \(y=(x + 1)^{2}(x - 1)(x - 3)\) is obtained by multiplying the leading terms of each factor.
\((x+1)^{2}=x^{2}+2x + 1\), and \((x^{2}+2x + 1)(x-1)(x - 3)=(x^{2}+2x + 1)(x^{2}-4x + 3)=x^{4}-4x^{3}+3x^{2}+2x^{3}-8x^{2}+6x+x^{2}-4x + 3=x^{4}-2x^{3}-4x^{2}+2x + 3\).
As \(x
ightarrow\pm\infty\), \(y = f(x)\approx x^{4}\). So, as \(x
ightarrow-\infty\), \(y
ightarrow+\infty\) and as \(x
ightarrow+\infty\), \(y
ightarrow+\infty\).

Step3: Analyze the behavior at the x - intercepts

• For \(x=-1\) (multiplicity \(m = 2\), even): The graph touches the x - axis at \(x=-1\). The derivative \(y^\prime\) (using the product rule \(y=(x + 1)^{2}(x - 1)(x - 3)\), \(y^\prime=2(x + 1)(x - 1)(x - 3)+(x + 1)^{2}(x - 3)+(x + 1)^{2}(x - 1)\). At \(x=-1\), \(y^\prime=0\).
• For \(x = 1\) (multiplicity \(m = 1\), odd): The graph crosses the x - axis at \(x = 1\).
• For \(x = 3\) (multiplicity \(m = 1\), odd): The graph crosses the x - axis at \(x = 3\).

Step4: Find the y - intercept

Set \(x = 0\). Then \(y=(0 + 1)^{2}(0 - 1)(0 - 3)=(1)\times(-1)\times(-3)=3\). So the y - intercept is \((0,3)\).

Answer:

Plot the x - intercepts \((-1,0)\), \((1,0)\), \((3,0)\), the y - intercept \((0,3)\). Use the end - behavior (as \(x
ightarrow\pm\infty\), \(y
ightarrow+\infty\)) and the behavior at the x - intercepts (touch at \(x=-1\), cross at \(x = 1\) and \(x = 3\)) to sketch the graph of \(y=(x + 1)^{2}(x - 1)(x - 3)\).