Question

determine an equation for the pictured graph. write your answer in factored form and assume the leading coefficient is either 1 or (-1), you should be able to determine which is the case by looking at the graph.
y =

Explanation:

Step1: Identify the roots and their multiplicities

From the graph, we can see that the graph crosses the x - axis at \(x=-2\) (a single root, so multiplicity 1), touches the x - axis at \(x = 0\) (a root with even multiplicity, let's assume multiplicity 2 for now) and crosses the x - axis at \(x = 1\) (a single root, multiplicity 1).

The general form of a polynomial in factored form is \(y=a(x - r_1)^{m_1}(x - r_2)^{m_2}(x - r_3)^{m_3}\), where \(r_i\) are the roots and \(m_i\) are their multiplicities.

So, with roots \(r_1=-2\), \(r_2 = 0\), \(r_3=1\) and multiplicities \(m_1 = 1\), \(m_2=2\), \(m_3 = 1\), the polynomial is \(y=a(x + 2)x^{2}(x - 1)\).

Step2: Determine the leading coefficient \(a\)

We need to determine if \(a = 1\) or \(a=-1\). Let's look at the end - behavior or a test point. Let's consider the degree of the polynomial. The degree of the polynomial \(y=a(x + 2)x^{2}(x - 1)=a(x^{4}+x^{3}-2x^{2})\) is 4 (even). For a polynomial of even degree, as \(x\to\pm\infty\), the leading term \(ax^{4}\) dominates.

Looking at the graph, as \(x\to+\infty\), the graph goes down (since the right - hand end of the graph is decreasing). For a polynomial \(y = ax^{n}\) with \(n\) even, if \(a<0\), as \(x\to\pm\infty\), \(y\to-\infty\). If \(a > 0\), as \(x\to\pm\infty\), \(y\to+\infty\). So, since the right - hand end is going down, \(a=-1\).

Step3: Write the final equation

Substitute \(a=-1\) into the factored form:

\(y=-1(x + 2)x^{2}(x - 1)=-x^{2}(x + 2)(x - 1)\)

We can also expand it a bit to check, but the factored form is what is required.

Answer:

\(y=-x^{2}(x + 2)(x - 1)\)