Math Problem Statement

What is the range of this quadratic function?

y

=

x

2

2

x

8

Solution

To find the range of the quadratic function y=x22x+8y = -x^2 - 2x + 8, let's go through the steps:

1. Identify the vertex of the parabola:

The function is in the form y=ax2+bx+cy = ax^2 + bx + c, where a=1a = -1, b=2b = -2, and c=8c = 8.

The x-coordinate of the vertex is calculated using the formula: x=b2ax = \frac{-b}{2a} Substituting the values of aa and bb: x=(2)2(1)=22=1x = \frac{-(-2)}{2(-1)} = \frac{2}{-2} = -1

2. Find the y-coordinate of the vertex:

Substitute x=1x = -1 back into the equation to find yy: y=(1)22(1)+8y = -(-1)^2 - 2(-1) + 8 y=1+2+8=9y = -1 + 2 + 8 = 9

3. Determine the range:

Since the coefficient of x2x^2 (which is a=1a = -1) is negative, the parabola opens downward. This means the vertex is the maximum point.

Therefore, the range of the function is: (,9](-\infty, 9]

Would you like more details or have any questions?

5 Related Questions:

  1. How do you find the x-intercepts of a quadratic function?
  2. What is the significance of the vertex in determining the maximum or minimum value?
  3. How can you convert a quadratic function to vertex form?
  4. How does the value of cc in the quadratic function affect the graph?
  5. What is the relationship between the axis of symmetry and the vertex?

Tip:

The range of a quadratic function depends on whether the parabola opens upward or downward. The vertex provides the highest or lowest point, which helps determine the range.

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Math Problem Analysis

Mathematical Concepts

Quadratic Functions
Vertex of a Parabola
Range of Functions

Formulas

Vertex formula: x = -b/(2a)
Quadratic function: y = ax^2 + bx + c

Theorems

Maximum/Minimum Value Theorem for Quadratic Functions

Suitable Grade Level

Grades 10-12