Answer :
To determine the range of the function [tex]\( f(x) = (x - 4)(x - 2) \)[/tex], we need to analyze its properties as a quadratic function. Let's go through the steps:
1. Rewrite the function in standard form:
First, expand the given function:
[tex]\[ f(x) = (x - 4)(x - 2) = x^2 - 2x - 4x + 8 = x^2 - 6x + 8 \][/tex]
Thus, the function is [tex]\( f(x) = x^2 - 6x + 8 \)[/tex].
2. Identify the shape and direction of the parabola:
The quadratic function [tex]\( f(x) = x^2 - 6x + 8 \)[/tex] is a parabola that opens upwards because the coefficient of [tex]\( x^2 \)[/tex] (a = 1) is positive.
3. Find the vertex of the parabola:
The vertex form of a parabolic function [tex]\( ax^2 + bx + c \)[/tex] gives the x-coordinate of the vertex as [tex]\( \frac{-b}{2a} \)[/tex]. Here, [tex]\( a = 1 \)[/tex] and [tex]\( b = -6 \)[/tex], so:
[tex]\[ x_{\text{vertex}} = \frac{-(-6)}{2(1)} = \frac{6}{2} = 3 \][/tex]
4. Calculate the y-coordinate of the vertex:
Substitute [tex]\( x = 3 \)[/tex] back into the function [tex]\( f(x) \)[/tex]:
[tex]\[ f(3) = (3 - 4)(3 - 2) = (-1)(1) = -1 \][/tex]
Therefore, the vertex of the parabola is at [tex]\( (3, -1) \)[/tex].
5. Determine the range of the function:
Since the parabola opens upwards, the vertex represents the minimum value of the function. Consequently, the y-coordinate of the vertex is the minimum value of [tex]\( f(x) \)[/tex]. Thus, the range of [tex]\( f(x) \)[/tex] is all real numbers greater than or equal to -1.
Therefore, the correct answer is:
[tex]\[ \text{The range of the function is } \text{all real numbers greater than or equal to -1.} \][/tex]
1. Rewrite the function in standard form:
First, expand the given function:
[tex]\[ f(x) = (x - 4)(x - 2) = x^2 - 2x - 4x + 8 = x^2 - 6x + 8 \][/tex]
Thus, the function is [tex]\( f(x) = x^2 - 6x + 8 \)[/tex].
2. Identify the shape and direction of the parabola:
The quadratic function [tex]\( f(x) = x^2 - 6x + 8 \)[/tex] is a parabola that opens upwards because the coefficient of [tex]\( x^2 \)[/tex] (a = 1) is positive.
3. Find the vertex of the parabola:
The vertex form of a parabolic function [tex]\( ax^2 + bx + c \)[/tex] gives the x-coordinate of the vertex as [tex]\( \frac{-b}{2a} \)[/tex]. Here, [tex]\( a = 1 \)[/tex] and [tex]\( b = -6 \)[/tex], so:
[tex]\[ x_{\text{vertex}} = \frac{-(-6)}{2(1)} = \frac{6}{2} = 3 \][/tex]
4. Calculate the y-coordinate of the vertex:
Substitute [tex]\( x = 3 \)[/tex] back into the function [tex]\( f(x) \)[/tex]:
[tex]\[ f(3) = (3 - 4)(3 - 2) = (-1)(1) = -1 \][/tex]
Therefore, the vertex of the parabola is at [tex]\( (3, -1) \)[/tex].
5. Determine the range of the function:
Since the parabola opens upwards, the vertex represents the minimum value of the function. Consequently, the y-coordinate of the vertex is the minimum value of [tex]\( f(x) \)[/tex]. Thus, the range of [tex]\( f(x) \)[/tex] is all real numbers greater than or equal to -1.
Therefore, the correct answer is:
[tex]\[ \text{The range of the function is } \text{all real numbers greater than or equal to -1.} \][/tex]
