Answer :
To find the vertex of the quadratic function [tex]\( a(x) = (x-7)(x+3) + 25 \)[/tex], we will follow these steps:
### Step 1: Expand the Function
First, we need to expand the function to get it into the standard quadratic form [tex]\( ax^2 + bx + c \)[/tex].
[tex]\[ a(x) = (x-7)(x+3) + 25 \][/tex]
Let's expand [tex]\((x-7)(x+3)\)[/tex]:
[tex]\[ = x^2 + 3x - 7x - 21 \][/tex]
[tex]\[ = x^2 - 4x - 21 \][/tex]
Now we add the constant term 25:
[tex]\[ a(x) = x^2 - 4x - 21 + 25 \][/tex]
[tex]\[ a(x) = x^2 - 4x + 4 \][/tex]
### Step 2: Find the First Derivative
To find the vertex, we need to determine where the first derivative of the function is zero, since the vertex of a quadratic function occurs at the maximum or minimum point.
The first derivative of [tex]\( a(x) = x^2 - 4x + 4 \)[/tex] is:
[tex]\[ a'(x) = 2x - 4 \][/tex]
### Step 3: Solve for the Vertex
Set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 2x - 4 = 0 \][/tex]
[tex]\[ 2x = 4 \][/tex]
[tex]\[ x = 2 \][/tex]
So, the x-coordinate of the vertex is [tex]\( x = 2 \)[/tex].
### Step 4: Find the y-coordinate of the Vertex
Substitute [tex]\( x = 2 \)[/tex] back into the original function [tex]\( a(x) \)[/tex] to find the y-coordinate:
[tex]\[ a(2) = (2-7)(2+3) + 25 \][/tex]
[tex]\[ a(2) = (-5)(5) + 25 \][/tex]
[tex]\[ a(2) = -25 + 25 \][/tex]
[tex]\[ a(2) = 0 \][/tex]
Therefore, the y-coordinate of the vertex is [tex]\( y = 0 \)[/tex].
### Conclusion
The vertex of the function [tex]\( a(x) = (x-7)(x+3) + 25 \)[/tex] is at the point [tex]\( (2, 0) \)[/tex].
### Step 1: Expand the Function
First, we need to expand the function to get it into the standard quadratic form [tex]\( ax^2 + bx + c \)[/tex].
[tex]\[ a(x) = (x-7)(x+3) + 25 \][/tex]
Let's expand [tex]\((x-7)(x+3)\)[/tex]:
[tex]\[ = x^2 + 3x - 7x - 21 \][/tex]
[tex]\[ = x^2 - 4x - 21 \][/tex]
Now we add the constant term 25:
[tex]\[ a(x) = x^2 - 4x - 21 + 25 \][/tex]
[tex]\[ a(x) = x^2 - 4x + 4 \][/tex]
### Step 2: Find the First Derivative
To find the vertex, we need to determine where the first derivative of the function is zero, since the vertex of a quadratic function occurs at the maximum or minimum point.
The first derivative of [tex]\( a(x) = x^2 - 4x + 4 \)[/tex] is:
[tex]\[ a'(x) = 2x - 4 \][/tex]
### Step 3: Solve for the Vertex
Set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 2x - 4 = 0 \][/tex]
[tex]\[ 2x = 4 \][/tex]
[tex]\[ x = 2 \][/tex]
So, the x-coordinate of the vertex is [tex]\( x = 2 \)[/tex].
### Step 4: Find the y-coordinate of the Vertex
Substitute [tex]\( x = 2 \)[/tex] back into the original function [tex]\( a(x) \)[/tex] to find the y-coordinate:
[tex]\[ a(2) = (2-7)(2+3) + 25 \][/tex]
[tex]\[ a(2) = (-5)(5) + 25 \][/tex]
[tex]\[ a(2) = -25 + 25 \][/tex]
[tex]\[ a(2) = 0 \][/tex]
Therefore, the y-coordinate of the vertex is [tex]\( y = 0 \)[/tex].
### Conclusion
The vertex of the function [tex]\( a(x) = (x-7)(x+3) + 25 \)[/tex] is at the point [tex]\( (2, 0) \)[/tex].
