Sure, let's go through the detailed step-by-step solution for determining the domain and range of the function [tex]\( f(x) = (x+2)(x+6) \)[/tex].
1. Identify the quadratic function:
The given function is [tex]\( f(x) = (x+2)(x+6) \)[/tex].
2. Expand the function to standard form:
Expanding the expression, we get:
[tex]\[
f(x) = (x+2)(x+6) = x^2 + 6x + 2x + 12 = x^2 + 8x + 12.
\][/tex]
3. Find the domain of the function:
For quadratic functions, the domain is all real numbers, because there are no restrictions on the values [tex]\( x \)[/tex] can take. In interval notation, this is [tex]\( (-\infty, \infty) \)[/tex].
4. Find the range of the function:
To find the range, we need to find the vertex of the parabola since it opens upwards (coefficient of [tex]\( x^2 \)[/tex] is positive).
The vertex form of a parabola [tex]\( ax^2 + bx + c \)[/tex] is found using the formula for the x-coordinate of the vertex:
[tex]\[
x = -\frac{b}{2a}.
\][/tex]
For the given quadratic function [tex]\( x^2 + 8x + 12 \)[/tex]:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = 8 \)[/tex]
The x-coordinate of the vertex is:
[tex]\[
x = -\frac{8}{2 \cdot 1} = -4.
\][/tex]
Substitute [tex]\( x = -4 \)[/tex] back into the function to find the y-coordinate of the vertex:
[tex]\[
f(-4) = (-4)^2 + 8(-4) + 12 = 16 - 32 + 12 = -4.
\][/tex]
Thus, the vertex of the parabola is at (-4, -4).
Since the parabola opens upwards, the range is all real numbers greater than or equal to the y-coordinate of the vertex. Hence, the range is:
[tex]\[
\text{All real numbers} \geq -4.
\][/tex]
5. Conclusion:
The domain of the function [tex]\( f(x) = (x+2)(x+6) \)[/tex] is all real numbers, and the range is all real numbers greater than or equal to -4.
So, the correct statement is:
The domain is all real numbers, and the range is all real numbers greater than or equal to -4.
