Answer :
To determine the range of the function [tex]\( f(x) = -(x+5)(x+1) \)[/tex], follow these steps:
1. Identify the Type of Function:
The given function is a quadratic function (a polynomial of degree 2).
2. Identify the Leading Coefficient:
From the given function, we can observe that the quadratic term will open downwards because of the negative sign in front.
3. Convert Function to Standard Form:
Expand the expression:
[tex]\[ f(x) = -(x+5)(x+1) = -[x^2 + 6x + 5] = -x^2 - 6x - 5 \][/tex]
4. Determine the Vertex Form:
The standard form of a parabola is [tex]\( f(x) = ax^2 + bx + c \)[/tex], where [tex]\( a = -1 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = -5 \)[/tex].
5. Find the Vertex:
The x-coordinate of the vertex for a parabola given by [tex]\( f(x) = ax^2 + bx + c \)[/tex] is found using [tex]\( x = -\frac{b}{2a} \)[/tex].
Given [tex]\( a = -1 \)[/tex] and [tex]\( b = -6 \)[/tex]:
[tex]\[ x = -\frac{-6}{2(-1)} = \frac{6}{-2} = -3 \][/tex]
6. Calculate the Maximum Value (y-coordinate of the Vertex):
Substitute [tex]\( x = -3 \)[/tex] back into the original function to find the y-coordinate:
[tex]\[ f(-3) = -((-3+5)(-3+1)) = -((2)(-2)) = -(-4) = 4 \][/tex]
7. Determine the Range:
Since the parabola opens downward and has its vertex at [tex]\( (-3, 4) \)[/tex], the maximum value occurs at [tex]\( y = 4 \)[/tex]. Therefore, the function takes all values less than or equal to this maximum value.
Thus, the range of the function [tex]\( f(x) = -(x+5)(x+1) \)[/tex] is:
[tex]\[ \boxed{\text{all real numbers less than or equal to 4}} \][/tex]
1. Identify the Type of Function:
The given function is a quadratic function (a polynomial of degree 2).
2. Identify the Leading Coefficient:
From the given function, we can observe that the quadratic term will open downwards because of the negative sign in front.
3. Convert Function to Standard Form:
Expand the expression:
[tex]\[ f(x) = -(x+5)(x+1) = -[x^2 + 6x + 5] = -x^2 - 6x - 5 \][/tex]
4. Determine the Vertex Form:
The standard form of a parabola is [tex]\( f(x) = ax^2 + bx + c \)[/tex], where [tex]\( a = -1 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = -5 \)[/tex].
5. Find the Vertex:
The x-coordinate of the vertex for a parabola given by [tex]\( f(x) = ax^2 + bx + c \)[/tex] is found using [tex]\( x = -\frac{b}{2a} \)[/tex].
Given [tex]\( a = -1 \)[/tex] and [tex]\( b = -6 \)[/tex]:
[tex]\[ x = -\frac{-6}{2(-1)} = \frac{6}{-2} = -3 \][/tex]
6. Calculate the Maximum Value (y-coordinate of the Vertex):
Substitute [tex]\( x = -3 \)[/tex] back into the original function to find the y-coordinate:
[tex]\[ f(-3) = -((-3+5)(-3+1)) = -((2)(-2)) = -(-4) = 4 \][/tex]
7. Determine the Range:
Since the parabola opens downward and has its vertex at [tex]\( (-3, 4) \)[/tex], the maximum value occurs at [tex]\( y = 4 \)[/tex]. Therefore, the function takes all values less than or equal to this maximum value.
Thus, the range of the function [tex]\( f(x) = -(x+5)(x+1) \)[/tex] is:
[tex]\[ \boxed{\text{all real numbers less than or equal to 4}} \][/tex]
