Math  /  Algebra

QuestionWrite an equation for the polynomial graphed below, in factored form. f(x)=f(x)= \square

Studdy Solution

STEP 1

What is this asking? We need to find the equation of a polynomial, written in factored form, based on its graph. Watch out! The graph *touches* the x-axis at the intercepts, it doesn't cross, so this means something special about the factors!

STEP 2

1. Identify the roots.
2. Determine the multiplicity of each root.
3. Construct the polynomial.
4. Determine the leading coefficient.

STEP 3

Looking at the graph, our polynomial has x-intercepts at x=3x = -3, x=1x = 1, and x=3x = 3.
These are the **roots** of our polynomial!

STEP 4

Notice how the graph just *touches* the x-axis at each root and bounces back?
That means each root has an **even multiplicity**.
The simplest even multiplicity is **2**.
Let's use that!

STEP 5

Since we know the roots and their multiplicities, we can start building our polynomial in factored form.
A root of x=3x = -3 with multiplicity 22 gives us a factor of (x(3))2=(x+3)2(x - (-3))^2 = (x+3)^2.

STEP 6

A root of x=1x = 1 with multiplicity 22 gives us a factor of (x1)2(x - 1)^2.

STEP 7

A root of x=3x = 3 with multiplicity 22 gives us a factor of (x3)2(x - 3)^2.

STEP 8

Putting it all together, our polynomial looks like f(x)=a(x+3)2(x1)2(x3)2f(x) = a(x+3)^2(x-1)^2(x-3)^2, where aa is some **leading coefficient** that we still need to figure out.

STEP 9

We know the graph opens upwards, so our leading coefficient aa must be **positive**.

STEP 10

We don't have any other specific points on the graph to help us determine the exact value of aa.
Since the problem doesn't ask for a specific value, and the general shape matches the graph, we can assume a=1a = 1.

STEP 11

So, our polynomial is f(x)=(x+3)2(x1)2(x3)2f(x) = (x+3)^2(x-1)^2(x-3)^2.

STEP 12

f(x)=(x+3)2(x1)2(x3)2f(x) = (x+3)^2(x-1)^2(x-3)^2

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