Question

write the equation of the polynomial shown below in standard form. be sure to show all of your steps for the show work problems. you may type your answers in the answer box using σ or upload a photo of your handwritten work using 📎 . if neither of these options work for you, send pictures of your work to your teacher in a webmail. if you do not show work to justify your answer, you will not receive credit. (4 points)

Explanation:

Step1: Identify roots from graph

The graph crosses the x - axis at \(x = 0\) (with a single root, so multiplicity 1) and touches the x - axis at \(x = 3\) (so multiplicity 2). Also, since the leading coefficient is positive (the graph goes up on the right) and the degree is \(1 + 2=3\)? Wait, no, the graph has a local maximum and a local minimum, and the end - behavior: as \(x
ightarrow+\infty\), \(y
ightarrow+\infty\) and as \(x
ightarrow-\infty\), \(y
ightarrow-\infty\), so the degree is odd. Wait, actually, the roots: let's assume the roots are \(x = 0\) (multiplicity 1) and \(x = 3\) (multiplicity 2). So the factored form is \(y=a(x - 0)(x - 3)^2=a x(x^{2}-6x + 9)=a(x^{3}-6x^{2}+9x)\). Now, we need to find the value of \(a\). Let's assume a point on the graph. Let's say when \(x = 1\), let's estimate the y - value. If we assume \(a = 1\) (since the graph seems to have a simple leading coefficient, but maybe we can check. Wait, maybe the graph is \(y=x(x - 3)^2\). Let's expand it: \(y=x(x^{2}-6x + 9)=x^{3}-6x^{2}+9x\). Wait, but let's check the end - behavior. The degree is 3, leading coefficient positive, so as \(x
ightarrow+\infty\), \(y
ightarrow+\infty\) and \(x
ightarrow-\infty\), \(y
ightarrow-\infty\), which matches the graph.

Step2: Expand the factored form

We have the factored form \(y = x(x - 3)^2\). First, expand \((x - 3)^2=x^{2}-6x + 9\). Then multiply by \(x\): \(y=x\times(x^{2}-6x + 9)=x^{3}-6x^{2}+9x\).

Answer:

\(y = x^{3}-6x^{2}+9x\)