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To sketch the graph of the polynomial function [tex]\( p(x) = (3x - 9)(x - 1) \)[/tex], let's break the process down into several steps.
### Step 1: Expand the Polynomial
Firstly, it's useful to expand the polynomial to better understand its components.
[tex]\[ p(x) = (3x - 9)(x - 1) \][/tex]
Using the distributive property, we expand it:
[tex]\[ p(x) = (3x \cdot x) + (3x \cdot -1) + (-9 \cdot x) + (-9 \cdot -1) \][/tex]
[tex]\[ p(x) = 3x^2 - 3x - 9x + 9 \][/tex]
Combining like terms, we get:
[tex]\[ p(x) = 3x^2 - 12x + 9 \][/tex]
### Step 2: Find the [tex]\(x\)[/tex]-Intercepts
The [tex]\( x \)[/tex]-intercepts occur where [tex]\( p(x) = 0 \)[/tex].
Setting the factored form to zero:
[tex]\[ (3x - 9)(x - 1) = 0 \][/tex]
This equation is satisfied when either [tex]\(3x - 9 = 0\)[/tex] or [tex]\(x - 1 = 0\)[/tex].
Solving for [tex]\(x\)[/tex]:
[tex]\[ 3x - 9 = 0 \implies x = 3 \][/tex]
[tex]\[ x - 1 = 0 \implies x = 1 \][/tex]
Thus, the [tex]\( x \)[/tex]-intercepts are at [tex]\( x = 3 \)[/tex] and [tex]\( x = 1 \)[/tex], giving us the points [tex]\((3, 0)\)[/tex] and [tex]\((1, 0)\)[/tex].
### Step 3: Find the [tex]\(y\)[/tex]-Intercept
The [tex]\(y\)[/tex]-intercept is found by evaluating [tex]\( p(x) \)[/tex] at [tex]\( x = 0 \)[/tex].
[tex]\[ p(0) = (3 \cdot 0 - 9)(0 - 1) = (-9)(-1) = 9 \][/tex]
So the [tex]\( y \)[/tex]-intercept is at [tex]\( (0, 9) \)[/tex].
### Step 4: Find the Vertex
The vertex of a parabola [tex]\( y = ax^2 + bx + c \)[/tex] is given by the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
For the polynomial [tex]\( 3x^2 - 12x + 9 \)[/tex], we have [tex]\( a = 3 \)[/tex] and [tex]\( b = -12 \)[/tex].
[tex]\[ x_{\text{vertex}} = -\frac{-12}{2 \cdot 3} = \frac{12}{6} = 2 \][/tex]
Now, we substitute [tex]\( x = 2 \)[/tex] back into [tex]\( p(x) \)[/tex] to find the [tex]\( y \)[/tex]-coordinate of the vertex.
[tex]\[ p(2) = 3(2)^2 - 12(2) + 9 = 3 \cdot 4 - 24 + 9 = 12 - 24 + 9 = -3 \][/tex]
So the vertex is at [tex]\( (2, -3) \)[/tex].
### Summary of Key Points
- The [tex]\( x \)[/tex]-intercepts are at [tex]\( (1, 0) \)[/tex] and [tex]\( (3, 0) \)[/tex].
- The [tex]\( y \)[/tex]-intercept is at [tex]\( (0, 9) \)[/tex].
- The vertex is at [tex]\( (2, -3) \)[/tex].
### Sketching the Graph
1. Plot the [tex]\( x \)[/tex]-intercepts [tex]\((1, 0)\)[/tex] and [tex]\((3, 0)\)[/tex].
2. Plot the [tex]\( y \)[/tex]-intercept [tex]\((0, 9)\)[/tex].
3. Plot the vertex [tex]\((2, -3)\)[/tex].
4. Draw the parabola through these points, noting that since [tex]\( a = 3 \)[/tex] (which is positive), the parabola opens upward.
Your sketch will show an upward-opening parabola passing through the critical points, providing a clear visual representation of the polynomial function.
### Step 1: Expand the Polynomial
Firstly, it's useful to expand the polynomial to better understand its components.
[tex]\[ p(x) = (3x - 9)(x - 1) \][/tex]
Using the distributive property, we expand it:
[tex]\[ p(x) = (3x \cdot x) + (3x \cdot -1) + (-9 \cdot x) + (-9 \cdot -1) \][/tex]
[tex]\[ p(x) = 3x^2 - 3x - 9x + 9 \][/tex]
Combining like terms, we get:
[tex]\[ p(x) = 3x^2 - 12x + 9 \][/tex]
### Step 2: Find the [tex]\(x\)[/tex]-Intercepts
The [tex]\( x \)[/tex]-intercepts occur where [tex]\( p(x) = 0 \)[/tex].
Setting the factored form to zero:
[tex]\[ (3x - 9)(x - 1) = 0 \][/tex]
This equation is satisfied when either [tex]\(3x - 9 = 0\)[/tex] or [tex]\(x - 1 = 0\)[/tex].
Solving for [tex]\(x\)[/tex]:
[tex]\[ 3x - 9 = 0 \implies x = 3 \][/tex]
[tex]\[ x - 1 = 0 \implies x = 1 \][/tex]
Thus, the [tex]\( x \)[/tex]-intercepts are at [tex]\( x = 3 \)[/tex] and [tex]\( x = 1 \)[/tex], giving us the points [tex]\((3, 0)\)[/tex] and [tex]\((1, 0)\)[/tex].
### Step 3: Find the [tex]\(y\)[/tex]-Intercept
The [tex]\(y\)[/tex]-intercept is found by evaluating [tex]\( p(x) \)[/tex] at [tex]\( x = 0 \)[/tex].
[tex]\[ p(0) = (3 \cdot 0 - 9)(0 - 1) = (-9)(-1) = 9 \][/tex]
So the [tex]\( y \)[/tex]-intercept is at [tex]\( (0, 9) \)[/tex].
### Step 4: Find the Vertex
The vertex of a parabola [tex]\( y = ax^2 + bx + c \)[/tex] is given by the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
For the polynomial [tex]\( 3x^2 - 12x + 9 \)[/tex], we have [tex]\( a = 3 \)[/tex] and [tex]\( b = -12 \)[/tex].
[tex]\[ x_{\text{vertex}} = -\frac{-12}{2 \cdot 3} = \frac{12}{6} = 2 \][/tex]
Now, we substitute [tex]\( x = 2 \)[/tex] back into [tex]\( p(x) \)[/tex] to find the [tex]\( y \)[/tex]-coordinate of the vertex.
[tex]\[ p(2) = 3(2)^2 - 12(2) + 9 = 3 \cdot 4 - 24 + 9 = 12 - 24 + 9 = -3 \][/tex]
So the vertex is at [tex]\( (2, -3) \)[/tex].
### Summary of Key Points
- The [tex]\( x \)[/tex]-intercepts are at [tex]\( (1, 0) \)[/tex] and [tex]\( (3, 0) \)[/tex].
- The [tex]\( y \)[/tex]-intercept is at [tex]\( (0, 9) \)[/tex].
- The vertex is at [tex]\( (2, -3) \)[/tex].
### Sketching the Graph
1. Plot the [tex]\( x \)[/tex]-intercepts [tex]\((1, 0)\)[/tex] and [tex]\((3, 0)\)[/tex].
2. Plot the [tex]\( y \)[/tex]-intercept [tex]\((0, 9)\)[/tex].
3. Plot the vertex [tex]\((2, -3)\)[/tex].
4. Draw the parabola through these points, noting that since [tex]\( a = 3 \)[/tex] (which is positive), the parabola opens upward.
Your sketch will show an upward-opening parabola passing through the critical points, providing a clear visual representation of the polynomial function.
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