2. REASON QUANTITATIVELY Explain why the $$ x $$-coordinates of the points of intersection of $$ f(x)=x^{3}+x^{2}-14 x-4 $$ and $$ g(x)=x^{3}-3 x^{2}-6 x+28 $$ represent the solutions to $$ x^{3}+x^{2}-14 x-4=x^{3}-3 x^{2}-6 x+28 $$. Then use the graphs to solve the equation.

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Certainly! Let's break down the problem step by step to understand why the $$ x $$-coordinates of the intersection points of the functions $$ f(x) $$ and $$ g(x) $$ represent the solutions to the equation $$ f(x) = g(x) $$, and then solve the equation using this reasoning.

## Understanding the Relationship Between Intersection Points and Solutions

### Quantitative Reasoning

1. **Intersection Points Definition**:
   - The **points of intersection** between two graphs $$ f(x) $$ and $$ g(x) $$ are the points where both functions have the same $$ y $$-value for the same $$ x $$-value. In other words, at these points, $$ f(x) = g(x) $$.

2. **Setting Up the Equation**:
   - To find the $$ x $$-coordinates where $$ f(x) $$ and $$ g(x) $$ intersect, we set the two functions equal to each other:
     $$
     f(x) = g(x)
     $$
     
3. **Solving for $$ x $$**:
   - By solving the equation $$ f(x) = g(x) $$, we determine the specific $$ x $$-values where the two functions yield the same output. These $$ x $$-values correspond to the horizontal positions of the intersection points on the graph.

Thus, **solving $$ f(x) = g(x) $$** directly gives us the $$ x $$-coordinates of the points where the graphs of $$ f(x) $$ and $$ g(x) $$ intersect.

## Solving the Equation Using Graphical Reasoning

Given the functions:
$$
f(x) = x^{3} + x^{2} - 14x - 4
$$
$$
g(x) = x^{3} - 3x^{2} - 6x + 28
$$

### Step 1: Set $$ f(x) $$ Equal to $$ g(x) $$
$$
x^{3} + x^{2} - 14x - 4 = x^{3} - 3x^{2} - 6x + 28
$$

### Step 2: Simplify the Equation
Subtract $$ x^{3} $$ from both sides to eliminate the cubic terms:
$$
x^{2} - 14x - 4 = -3x^{2} - 6x + 28
$$

### Step 3: Bring All Terms to One Side
Add $$ 3x^{2} + 6x - 28 $$ to both sides:
$$
x^{2} + 3x^{2} - 14x + 6x - 4 - 28 = 0
$$
$$
4x^{2} - 8x - 32 = 0
$$

### Step 4: Simplify the Quadratic Equation
Divide the entire equation by 4 to make it simpler:
$$
x^{2} - 2x - 8 = 0
$$

### Step 5: Solve the Quadratic Equation
Use the quadratic formula:
$$
x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}
$$
For the equation $$ x^{2} - 2x - 8 = 0 $$:
- $$ a = 1 $$
- $$ b = -2 $$
- $$ c = -8 $$

Plugging in the values:
$$
x = \frac{-(-2) \pm \sqrt{(-2)^{2} - 4(1)(-8)}}{2(1)}
$$
$$
x = \frac{2 \pm \sqrt{4 + 32}}{2}
$$
$$
x = \frac{2 \pm \sqrt{36}}{2}
$$
$$
x = \frac{2 \pm 6}{2}
$$

This gives two solutions:
$$
x = \frac{2 + 6}{2} = \frac{8}{2} = 4
$$
$$
x = \frac{2 - 6}{2} = \frac{-4}{2} = -2
$$

### Step 6: Interpret the Solutions
The solutions $$ x = 4 $$ and $$ x = -2 $$ are the $$ x $$-coordinates where the graphs of $$ f(x) $$ and $$ g(x) $$ intersect.

### Graphical Confirmation (Optional)

If you were to graph both functions:

- **At $$ x = 4 $$**:
  $$
  f(4) = 4^{3} + 4^{2} - 14(4) - 4 = 64 + 16 - 56 - 4 = 20
  $$
  $$
  g(4) = 4^{3} - 3(4)^{2} - 6(4) + 28 = 64 - 48 - 24 + 28 = 20
  $$
  
- **At $$ x = -2 $$**:
  $$
  f(-2) = (-2)^{3} + (-2)^{2} - 14(-2) - 4 = -8 + 4 + 28 - 4 = 20
  $$
  $$
  g(-2) = (-2)^{3} - 3(-2)^{2} - 6(-2) + 28 = -8 - 12 + 12 + 28 = 20
  $$

Both functions yield the same $$ y $$-value at $$ x = 4 $$ and $$ x = -2 $$, confirming these points as intersections.

## **Final Answer**

The solutions to the equation $$ f(x) = g(x) $$ are:
$$
\boxed{\,x = -2\quad\text{and}\quad x = 4\,}
$$
The solutions to the equation $$ x^{3} + x^{2} - 14x - 4 = x^{3} - 3x^{2} - 6x + 28 $$ are $$ x = -2 $$ and $$ x = 4 $$.

REASON QUANTITATIVELY Explain why the -coordinates of the points of
intersection of and
represent the solutions to . Then use
the graphs to solve the equation.

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REASON QUANTITATIVELY Explain why the -coordinates of the points of intersection of and represent the solutions to . Then use the graphs to solve the equation.