Answer :
Let's solve the quadratic function [tex]\( y = x^2 + 2x - 8 \)[/tex]. Here are the key features we need to determine: factored form, x-intercepts (roots), axis of symmetry, vertex, domain, and range.
### 1. Factored Form
To factor the quadratic function [tex]\( y = x^2 + 2x - 8 \)[/tex]:
[tex]\[ y = (x - 2)(x + 4) \][/tex]
### 2. x-Intercepts (Roots)
The x-intercepts are the values of [tex]\( x \)[/tex] that make [tex]\( y \)[/tex] equal to zero. We solve [tex]\( (x - 2)(x + 4) = 0 \)[/tex]:
[tex]\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \][/tex]
[tex]\[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \][/tex]
So the x-intercepts are:
[tex]\[ x = -4 \quad \text{and} \quad x = 2 \][/tex]
### 3. Axis of Symmetry
The axis of symmetry for a quadratic function [tex]\( ax^2 + bx + c \)[/tex] is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
For our function [tex]\( y = x^2 + 2x - 8 \)[/tex]:
[tex]\[ a = 1, \quad b = 2 \][/tex]
[tex]\[ x = -\frac{2}{2 \cdot 1} = -1 \][/tex]
So the axis of symmetry is:
[tex]\[ x = -1 \][/tex]
### 4. Vertex
The vertex is the point [tex]\((x, y)\)[/tex] on the curve where the y value is at its minimum (for a parabola that opens upwards).
Substituting the x value of the axis of symmetry into the original function to find the y value:
[tex]\[ y = (-1)^2 + 2(-1) - 8 \][/tex]
[tex]\[ y = 1 - 2 - 8 = -9 \][/tex]
So the vertex is:
[tex]\[ (-1, -9) \][/tex]
### 5. Domain
The domain of a quadratic function is all real numbers:
[tex]\[ \text{Domain}: (-\infty, \infty) \quad \text{or} \quad \mathbb{R} \][/tex]
### 6. Range
The range is determined by the vertex and the direction in which the parabola opens. Because this parabola opens upwards (the coefficient of [tex]\( x^2 \)[/tex] is positive), the vertex represents the minimum y-value.
Therefore, the range is:
[tex]\[ \text{Range}: [-9, \infty) \][/tex]
### Summary
- Factored Form: [tex]\( y = (x - 2)(x + 4) \)[/tex]
- x-Intercepts: [tex]\( x = -4 \text{ and } x = 2 \)[/tex]
- Axis of Symmetry: [tex]\( x = -1 \)[/tex]
- Vertex: [tex]\( (-1, -9) \)[/tex]
- Domain: [tex]\( \mathbb{R} \)[/tex] (all real numbers)
- Range: [tex]\( [-9, \infty) \)[/tex]
### 1. Factored Form
To factor the quadratic function [tex]\( y = x^2 + 2x - 8 \)[/tex]:
[tex]\[ y = (x - 2)(x + 4) \][/tex]
### 2. x-Intercepts (Roots)
The x-intercepts are the values of [tex]\( x \)[/tex] that make [tex]\( y \)[/tex] equal to zero. We solve [tex]\( (x - 2)(x + 4) = 0 \)[/tex]:
[tex]\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \][/tex]
[tex]\[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \][/tex]
So the x-intercepts are:
[tex]\[ x = -4 \quad \text{and} \quad x = 2 \][/tex]
### 3. Axis of Symmetry
The axis of symmetry for a quadratic function [tex]\( ax^2 + bx + c \)[/tex] is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
For our function [tex]\( y = x^2 + 2x - 8 \)[/tex]:
[tex]\[ a = 1, \quad b = 2 \][/tex]
[tex]\[ x = -\frac{2}{2 \cdot 1} = -1 \][/tex]
So the axis of symmetry is:
[tex]\[ x = -1 \][/tex]
### 4. Vertex
The vertex is the point [tex]\((x, y)\)[/tex] on the curve where the y value is at its minimum (for a parabola that opens upwards).
Substituting the x value of the axis of symmetry into the original function to find the y value:
[tex]\[ y = (-1)^2 + 2(-1) - 8 \][/tex]
[tex]\[ y = 1 - 2 - 8 = -9 \][/tex]
So the vertex is:
[tex]\[ (-1, -9) \][/tex]
### 5. Domain
The domain of a quadratic function is all real numbers:
[tex]\[ \text{Domain}: (-\infty, \infty) \quad \text{or} \quad \mathbb{R} \][/tex]
### 6. Range
The range is determined by the vertex and the direction in which the parabola opens. Because this parabola opens upwards (the coefficient of [tex]\( x^2 \)[/tex] is positive), the vertex represents the minimum y-value.
Therefore, the range is:
[tex]\[ \text{Range}: [-9, \infty) \][/tex]
### Summary
- Factored Form: [tex]\( y = (x - 2)(x + 4) \)[/tex]
- x-Intercepts: [tex]\( x = -4 \text{ and } x = 2 \)[/tex]
- Axis of Symmetry: [tex]\( x = -1 \)[/tex]
- Vertex: [tex]\( (-1, -9) \)[/tex]
- Domain: [tex]\( \mathbb{R} \)[/tex] (all real numbers)
- Range: [tex]\( [-9, \infty) \)[/tex]