Math Problem Statement

Use the graph to find the solution set of the equation -x^2 + 4x + 12 = -2x + 4.

Solution

The image contains a problem requiring the solution to an equation x2+4x+12=2x+4-x^2 + 4x + 12 = -2x + 4 using the graph.

Step-by-step solution:

  1. Simplify the given equation: x2+4x+12=2x+4-x^2 + 4x + 12 = -2x + 4 Move all terms to one side of the equation: x2+4x+2x+124=0-x^2 + 4x + 2x + 12 - 4 = 0 Simplify: x2+6x+8=0-x^2 + 6x + 8 = 0

  2. Identify the graph: From the image, the graph represents two functions:

    • The parabola y=x2+6x+8y = -x^2 + 6x + 8, and
    • The line y=2x+4y = -2x + 4.
  3. Find the intersection points: The solutions to x2+6x+8=2x+4-x^2 + 6x + 8 = -2x + 4 are the x-coordinates where the parabola intersects the line. From the graph, visually locate these intersection points.

    From the image, the approximate solutions appear to be: x=2andx=4x = 2 \quad \text{and} \quad x = 4

Final Answer:

x=2,4x = 2, 4

Would you like further clarification or assistance?


Here are five related questions to deepen your understanding:

  1. How can you verify the solutions algebraically?
  2. What does the discriminant of a quadratic equation tell us about its solutions?
  3. How do you interpret solutions graphically for systems of equations?
  4. How would the solution change if the quadratic and linear functions didn’t intersect?
  5. What are other methods to solve quadratic equations besides graphing?

Tip: Always check your solutions by substituting them back into the original equation.

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Math Problem Analysis

Mathematical Concepts

Graphing
Quadratic Equations
Systems of Equations

Formulas

Quadratic equation formula ax^2 + bx + c = 0

Theorems

Intersection points of a quadratic and a linear equation

Suitable Grade Level

Grades 8-10