Sure, let's analyze the function [tex]\( f(x) = (x + 2)(x + 6) \)[/tex] in detail to determine its domain and range.
1. Expanding the Function:
[tex]\[ f(x) = (x + 2)(x + 6) \][/tex]
[tex]\[ f(x) = x^2 + 6x + 2x + 12 \][/tex]
[tex]\[ f(x) = x^2 + 8x + 12 \][/tex]
2. Domain:
Since [tex]\( f(x) \)[/tex] is a quadratic function, it’s a polynomial, and polynomials are defined for all real numbers.
Hence, the domain of [tex]\( f(x) \)[/tex] is all real numbers.
3. Finding the Minimum Value (Range):
A quadratic function [tex]\( f(x) = ax^2 + bx + c \)[/tex] in the form [tex]\( ax^2 + bx + c \)[/tex] typically forms a parabola. In this case, the coefficient of [tex]\( x^2 \)[/tex] is positive ([tex]\( a = 1 \)[/tex]), so the parabola opens upwards. This means that the function has a minimum value, which occurs at the vertex of the parabola.
The x-coordinate of the vertex can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Plugging in the values [tex]\( a = 1 \)[/tex] and [tex]\( b = 8 \)[/tex]:
[tex]\[ x = -\frac{8}{2 \cdot 1} \][/tex]
[tex]\[ x = -4 \][/tex]
To find the corresponding y-coordinate (the minimum value of the function), substitute [tex]\( x = -4 \)[/tex] back into the function:
[tex]\[ f(-4) = (-4 + 2)(-4 + 6) \][/tex]
[tex]\[ f(-4) = (-2)(2) \][/tex]
[tex]\[ f(-4) = -4 \][/tex]
Therefore, the minimum value of the function is [tex]\(-4\)[/tex].
Since the parabola opens upwards, the range of the function starts at this minimum value and goes to infinity. Thus, the range of [tex]\( f(x) \)[/tex] is all real numbers greater than or equal to [tex]\(-4\)[/tex].
So, the correct statement is:
The domain is all real numbers, and the range is all real numbers greater than or equal to -4.
