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Math  /  Algebra

Questionf(x)=(x+1)2(x−4)3(x−3)f(x)=(x+1)^{2}(x-4)^{3}(x-3)
Find the zero(s) at which ff "flattens out". Express the zero(s) as ordered pair(s)

Studdy Solution

STEP 1

What is this asking? Find the points where this curvy function f(x)f(x) touches the x-axis and kinda hangs out there for a bit before crossing over or bouncing back. Watch out! Don't forget that "flattening out" means the function touches the x-axis *and* its slope is zero at that point!
This translates to a zero with a multiplicity greater than 1.

STEP 2

1. Find the zeros
2. Find their multiplicity
3. Express as ordered pairs

STEP 3

To find the zeros of f(x)f(x), we need to figure out the xx values that make the whole function equal zero.
Let's **set** f(x)f(x) **equal to zero**: (x+1)2(x−4)3(x−3)=0(x+1)^{2}(x-4)^{3}(x-3) = 0

STEP 4

Now, we have a product of factors equal to zero.
This means at least one of the factors *must* be zero.
So, we set each factor equal to zero and solve for xx: (x+1)2=0  ⟹  x+1=0  ⟹  x=−1(x+1)^2 = 0 \implies x+1 = 0 \implies x = -1 (x−4)3=0  ⟹  x−4=0  ⟹  x=4(x-4)^3 = 0 \implies x-4 = 0 \implies x = 4x−3=0  ⟹  x=3x-3 = 0 \implies x = 3So, our zeros are x=−1x = -1, x=4x = 4, and x=3x = 3.
Awesome!

STEP 5

The **multiplicity** of a zero tells us how many times that zero is a factor.
We can see the multiplicity by looking at the exponent of each factor.

STEP 6

For x=−1x = -1, the factor is (x+1)2(x+1)^2, and the exponent is **2**.
So, the multiplicity of x=−1x = -1 is **2**.
For x=4x = 4, the factor is (x−4)3(x-4)^3, and the exponent is **3**.
So, the multiplicity of x=4x = 4 is **3**.
For x=3x = 3, the factor is (x−3)(x-3), which can be thought of as (x−3)1(x-3)^1, and the exponent is **1**.
So, the multiplicity of x=3x = 3 is **1**.

STEP 7

We're looking for the zeros where the function "flattens out." This happens when the multiplicity of the zero is greater than 1.
In our case, that's x=−1x = -1 with multiplicity **2** and x=4x = 4 with multiplicity **3**.

STEP 8

Remember, an ordered pair is written as (x,f(x))(x, f(x)).
Since these are zeros, the f(x)f(x) value will be zero for both.
So, our ordered pairs are (−1,0)(-1, 0) and (4,0)(4, 0).
Boom!

STEP 9

The zeros at which ff "flattens out" are (−1,0)(-1, 0) and (4,0)(4, 0).

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