Question

1. on the grid below, sketch a cubic polynomial whose zeros are 1, 3, and -2.

Explanation:

Step1: Write the polynomial

Since the zeros are \(x = 1\), \(x=3\) and \(x = - 2\), the cubic polynomial can be written in factored - form as \(y=a(x - 1)(x - 3)(x + 2)\), where \(a
eq0\). For simplicity, let \(a = 1\), so \(y=(x - 1)(x - 3)(x + 2)\).

Step2: Expand the polynomial

\[

$$\begin{align*} y&=(x - 1)(x^{2}-3x + 2x-6)\\ &=(x - 1)(x^{2}-x - 6)\\ &=x^{3}-x^{2}-6x-x^{2}+x + 6\\ &=x^{3}-2x^{2}-5x + 6 \end{align*}$$

\]

Step3: Analyze the end - behavior

The leading coefficient of \(y=x^{3}-2x^{2}-5x + 6\) is \(1\) (positive) and the degree is \(3\) (odd). So as \(x\to-\infty\), \(y\to-\infty\) and as \(x\to+\infty\), \(y\to+\infty\).

Step4: Find the \(y\) - intercept

Set \(x = 0\) in \(y=x^{3}-2x^{2}-5x + 6\), then \(y=6\).

Step5: Sketch the graph

Plot the \(x\) - intercepts at \(x=-2\), \(x = 1\) and \(x = 3\), and the \(y\) - intercept at \((0,6)\). Use the end - behavior to sketch the curve passing through these points.

Answer:

Sketch a curve passing through the points \((-2,0)\), \((1,0)\), \((3,0)\) and \((0,6)\) with end - behavior: as \(x\to-\infty\), \(y\to-\infty\) and as \(x\to+\infty\), \(y\to+\infty\).