3 for each graph of a polynomial function, state the minimum degree, state if it has a positive or negative leading coefficient, and determine the domain and range
a)
4. the graph of a polynomial function ( y = p(x) ) is shown.
a) identify its factors and lowest possible multiplicities.
b) what other multiplicities are possible?
c) determine the zeros of each transformed function.
(i) ( f(x) = -\frac{1}{6}p(x) )
(iii) ( h(x) = p(x - 10) )
d) determine the ( y )-intercept of the graph for each transformed function.
(i) ( k(x) = p(x) - 20 )
(ii) ( m(x) = -\frac{3}{2}p(x) + 9 )

Question

Accepted Answer

### Part 4c(i): Zeros of $ f(x) = -\frac{1}{6}P(x) $
# Explanation:
## Step 1: Recall Zero Property
The zeros of a function $ f(x) = a \cdot g(x) $ (where $ a
eq 0 $) are the same as the zeros of $ g(x) $, since multiplying by a non - zero constant does not change the $ x $ - values where the function equals zero.
## Step 2: Analyze $ P(x) $'s Zeros
From part 4a, we know that the factors of $ P(x) $ are $ (x + 2)^3 $ and $ (x - 1)^2 $. The zeros of $ P(x) $ occur when $ (x + 2)^3=0 $ or $ (x - 1)^2 = 0 $. Solving $ (x + 2)^3=0 $ gives $ x=-2 $, and solving $ (x - 1)^2=0 $ gives $ x = 1 $.
## Step 3: Find Zeros of $ f(x) $
Since $ f(x)=-\frac{1}{6}P(x) $, the zeros of $ f(x) $ are the same as the zeros of $ P(x) $. So we set $ f(x) = 0 $, which implies $ -\frac{1}{6}P(x)=0 $, and then $ P(x)=0 $. The solutions to $ P(x) = 0 $ are $ x=-2 $ (with multiplicity 3) and $ x = 1 $ (with multiplicity 2).

# Answer:
The zeros of $ f(x)=-\frac{1}{6}P(x) $ are $ x=-2 $ (multiplicity 3) and $ x = 1 $ (multiplicity 2).

### Part 4c(iii): Zeros of $ h(x)=P(x - 10) $
# Explanation:
## Step 1: Recall Horizontal Shift Property
For a function $ y = P(x - h) $, it is a horizontal shift of the function $ y = P(x) $ by $ h $ units. The zeros of $ y = P(x - h) $ are found by setting $ x - h $ equal to the zeros of $ P(x) $.
## Step 2: Find Zeros of $ h(x) $
Let the zeros of $ P(x) $ be $ x=-2 $ and $ x = 1 $ (from part 4a). For $ h(x)=P(x - 10) $, we set $ x-10=-2 $ and $ x - 10=1 $.
- Solving $ x-10=-2 $: Add 10 to both sides of the equation, we get $ x=-2 + 10=8 $.
- Solving $ x - 10=1 $: Add 10 to both sides of the equation, we get $ x=1 + 10=11 $.

# Answer:
The zeros of $ h(x)=P(x - 10) $ are $ x = 8 $ (with the same multiplicity as the zero $ x=-2 $ of $ P(x) $, which is 3) and $ x = 11 $ (with the same multiplicity as the zero $ x = 1 $ of $ P(x) $, which is 2).

### Part 4d(i): $ y $ - intercept of $ k(x)=P(x)-20 $
# Explanation:
## Step 1: Recall $ y $ - intercept Definition
The $ y $ - intercept of a function $ y = k(x) $ is the value of the function when $ x = 0 $, i.e., $ k(0) $.
## Step 2: Analyze $ P(0) $
First, we need to find $ P(0) $. From the graph of $ P(x) $, we can see that when $ x = 0 $, we need to find the $ y $ - value of $ P(x) $. Let's assume from the graph (or from the fact that we can analyze the polynomial) that $ P(0) $ is the $ y $ - intercept of $ P(x) $. Looking at the graph, when $ x = 0 $, let's say $ P(0)=y_0 $. From the graph, we can estimate (or if we know the polynomial) that $ P(0) $ (the $ y $ - intercept of $ P(x) $): Let's assume that from the graph, when $ x = 0 $, the value of $ P(x) $ is, for example, if we consider the polynomial $ P(x)=(x + 2)^3(x - 1)^2 $, then $ P(0)=(0 + 2)^3(0 - 1)^2=(8)\times(1)=8 $. Wait, but maybe from the graph, let's re - check. Wait, the graph of $ P(x) $ has a $ y $ - intercept. Let's assume that when $ x = 0 $, $ P(0) $ is the $ y $ - value at $ x = 0 $. Let's say from the graph, the $ y $ - intercept of $ P(x) $ is $ P(0)=8 $ (we can also calculate it from the factored form: $ P(x)=(x + 2)^3(x - 1)^2 $, so $ P(0)=(2)^3(-1)^2=8\times1 = 8 $).
## Step 3: Calculate $ k(0) $
Now, $ k(x)=P(x)-20 $, so $ k(0)=P(0)-20 $. Substituting $ P(0) = 8 $, we get $ k(0)=8-20=-12 $.

# Answer:
The $ y $ - intercept of $ k(x)=P(x)-20 $ is $ - 12 $.

### Part 4d(ii): $ y $ - intercept of $ m(x)=-\frac{3}{2}P(x)+9 $
# Explanation:
## Step 1: Recall $ y $ - intercept Definition
The $ y $ - intercept of $ m(x) $ is $ m(0) $, which is the value of the function when $ x = 0 $.
## Step 2: Calculate $ P(0) $
As in part 4d(i), $ P(0)=(0 + 2)^3(0 - 1)^2=8\times1 = 8 $ (or from the graph).
## Step 3: Calculate $ m(0) $
Substitute $ x = 0 $ into $ m(x) $: $ m(0)=-\frac{3}{2}P(0)+9 $. Substitute $ P(0) = 8 $ into the equation: $ m(0)=-\frac{3}{2}\times8+9 $. First, calculate $ -\frac{3}{2}\times8=-12 $. Then, $ -12 + 9=-3 $.

# Answer:
The $ y $ - intercept of $ m(x)=-\frac{3}{2}P(x)+9 $ is $ -3 $.

### Part 4b: Other Possible Multiplicities
# Explanation:
## Step 1: Recall Multiplicity Rules
For a polynomial function, the multiplicity of a zero is a positive integer. If a zero has a multiplicity $ n $, then $ n\geq1 $, and for a zero from a factor $ (x - a)^n $, the graph touches the $ x $ - axis and turns around (local maximum or minimum) at $ x = a $ if $ n $ is even, and crosses the $ x $ - axis at $ x = a $ if $ n $ is odd.
## Step 2: Analyze Given Multiplicities
From part 4a, the lowest possible multiplicities are 3 for $ x=-2 $ (since the graph crosses the $ x $ - axis at $ x=-2 $, so multiplicity is odd, and the lowest odd positive integer is 1? Wait, no, wait the graph: if the graph crosses the $ x $ - axis at $ x=-2 $, the multiplicity should be odd, and if it touches and turns at $ x = 1 $, the multiplicity is even. Wait, maybe I made a mistake earlier. Wait, the graph of $ P(x) $: at $ x=-2 $, the graph crosses the $ x $ - axis, so the multiplicity of $ x=-2 $ is odd (1, 3, 5, ...), and at $ x = 1 $, the graph touches the $ x $ - axis and turns around, so the multiplicity of $ x = 1 $ is even (2, 4, 6, ...).
## Step 3: Determine Other Multiplicities
The lowest possible multiplicities are 1 for $ x=-2 $ (wait, no, maybe the initial analysis was wrong). Wait, let's re - examine. If the graph crosses the $ x $ - axis at $ x=-2 $, the multiplicity is odd (1, 3, 5,...), and touches at $ x = 1 $, multiplicity is even (2, 4, 6,...). So other possible multiplicities for $ x=-2 $ are $ 5,7,\cdots $ (any odd integer greater than 3) and for $ x = 1 $ are $ 4,6,8,\cdots $ (any even integer greater than 2). In general, for the zero $ x=-2 $ (where the graph crosses the $ x $ - axis), the multiplicity can be any odd positive integer ($ 1,3,5,\cdots $), and for the zero $ x = 1 $ (where the graph touches the $ x $ - axis and turns), the multiplicity can be any even positive integer ($ 2,4,6,\cdots $). So other possible multiplicities: for $ x=-2 $, multiplicities like 5, 7, etc., and for $ x = 1 $, multiplicities like 4, 6, etc. (as long as the multiplicity of $ x=-2 $ is odd and $ x = 1 $ is even, and the sum of multiplicities gives the degree of the polynomial).

# Answer:
For the zero $ x=-2 $ (where the graph crosses the $ x $ - axis), the multiplicity can be any odd positive integer greater than 1 (e.g., 3, 5, 7, ...) and for the zero $ x = 1 $ (where the graph touches the $ x $ - axis and turns), the multiplicity can be any even positive integer greater than 2 (e.g., 4, 6, 8, ...). In general, the multiplicity of $ x=-2 $ is an odd positive integer ($ 1,3,5,\cdots $) and the multiplicity of $ x = 1 $ is an even positive integer ($ 2,4,6,\cdots $), so other possible multiplicities include (but are not limited to) multiplicity 5 for $ x=-2 $ and multiplicity 4 for $ x = 1 $, multiplicity 7 for $ x=-2 $ and multiplicity 6 for $ x = 1 $, etc.

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QUESTION IMAGE

Question

Explanation:

Part 4c(i): Zeros of \( f(x) = -\frac{1}{6}P(x) \)

Step 1: Recall Zero Property

Step 2: Analyze \( P(x) \)'s Zeros

Step 3: Find Zeros of \( f(x) \)

Step 1: Recall Horizontal Shift Property

Step 2: Find Zeros of \( h(x) \)

Step 1: Recall \( y \) - intercept Definition

Step 2: Analyze \( P(0) \)

Step 3: Calculate \( k(0) \)

Answer:

Part 4c(iii): Zeros of \( h(x)=P(x - 10) \)