Certainly! Let's determine the vertex of the function [tex]\( f(x) = x^2 - 18x + 60 \)[/tex] step-by-step and analyze the statements provided.
### Step 1: Find the Vertex of the Parabola
The function given is a quadratic equation in the form [tex]\( f(x) = ax^2 + bx + c \)[/tex] where [tex]\( a = 1 \)[/tex], [tex]\( b = -18 \)[/tex], and [tex]\( c = 60 \)[/tex].
The vertex form of a quadratic equation can be derived using the formula for the x-coordinate of the vertex:
[tex]\[ x = \frac{-b}{2a} \][/tex]
Plugging in the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ x = \frac{-(-18)}{2 \cdot 1} = \frac{18}{2} = 9 \][/tex]
So, the x-coordinate of the vertex is [tex]\( x = 9 \)[/tex].
### Step 2: Calculate the y-coordinate of the Vertex
To find the y-coordinate of the vertex, substitute [tex]\( x = 9 \)[/tex] back into the function [tex]\( f(x) \)[/tex]:
[tex]\[ f(9) = (9)^2 - 18 \cdot 9 + 60 \][/tex]
[tex]\[ f(9) = 81 - 162 + 60 \][/tex]
[tex]\[ f(9) = -21 \][/tex]
So, the y-coordinate of the vertex is [tex]\( y = -21 \)[/tex].
### Step 3: Find the y-Intercept of the Function
The y-intercept of the function occurs when [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = (0)^2 - 18 \cdot 0 + 60 = 60 \][/tex]
So, the y-intercept is [tex]\( y = 60 \)[/tex].
### Step 4: Analyze the Statements
1. The x-coordinate of the vertex is greater than the y-coordinate.
[tex]\[ 9 > -21 \][/tex]
This statement is True.
2. The x-coordinate of the vertex is negative.
[tex]\[ 9 \][/tex] (The x-coordinate) is not negative.
This statement is False.
3. The y-coordinate of the vertex is greater than the y-intercept.
[tex]\[ -21 > 60 \][/tex]
This is False.
4. The y-coordinate of the vertex is positive.
[tex]\[ -21 \][/tex]
This is not positive.
This statement is False.
### Conclusion
Based on our detailed analysis, the statement that is true about the vertex of the function is:
The x-coordinate of the vertex is greater than the y-coordinate.
Therefore, the correct statement is:
The x-coordinate of the vertex is greater than the y-coordinate.
