Let's analyze the quadratic equation [tex]\(y = x^2 + 2x - 5\)[/tex] step-by-step to determine its domain and range.
### Step 1: Domain
For any quadratic function of the form [tex]\(y = ax^2 + bx + c\)[/tex], the domain is all real numbers. This means that the function is defined for every possible value of [tex]\(x\)[/tex]. So, we can state:
[tex]\[ \text{Domain: } D = \text{all real numbers} \][/tex]
### Step 2: Determine the Vertex
Next, we need to determine the range of the function. To do this, we first find the vertex of the quadratic function. The quadratic equation is in the standard form [tex]\(y = ax^2 + bx + c\)[/tex].
For [tex]\(y = x^2 + 2x - 5\)[/tex]:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = 2\)[/tex]
- [tex]\(c = -5\)[/tex]
The x-coordinate of the vertex ([tex]\(h\)[/tex]) can be found using the formula:
[tex]\[ h = -\frac{b}{2a} \][/tex]
Substituting the values:
[tex]\[ h = -\frac{2}{2 \cdot 1} = -1 \][/tex]
To find the y-coordinate of the vertex ([tex]\(k\)[/tex]), we substitute [tex]\(x = -1\)[/tex] back into the original equation:
[tex]\[ k = a(-1)^2 + b(-1) + c \][/tex]
[tex]\[ k = 1(-1)^2 + 2(-1) - 5 \][/tex]
[tex]\[ k = 1 - 2 - 5 \][/tex]
[tex]\[ k = -6 \][/tex]
So, the vertex of the parabola is [tex]\((-1, -6)\)[/tex].
### Step 3: Determine the Range
Since the parabola [tex]\(y = x^2 + 2x - 5\)[/tex] opens upwards (the coefficient of [tex]\(x^2\)[/tex] is positive), the y-coordinate of the vertex is the minimum value of the function. Thus, the range of the function is [tex]\(y \geq -6\)[/tex].
[tex]\[ \text{Range: } R = (y \geq -6) \][/tex]
### Step 4: Selecting the Best Answer
Given the domain and range:
- Domain: all real numbers
- Range: [tex]\(y \geq -6\)[/tex]
We can select the option that matches:
[tex]\[ \text{Option D:} \][/tex]
- Domain: all real numbers
- Range: [tex]\(y \geq -6\)[/tex]
Thus, the best answer is:
[tex]\[ \boxed{D} \][/tex]
