Step-by-step explanation:
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Given that \( f(x) \) is a quadratic function, we can express it in the form:
\[ f(x) = ax^2 + bx + c \]
We know two points on the graph of \( f(x) \): \( (1, 3) \) and \( (5, 3) \).
First, using \( f(1) = 3 \):
\[ a(1)^2 + b(1) + c = 3 \]
\[ a + b + c = 3 \]
Second, using \( f(5) = 3 \):
\[ a(5)^2 + b(5) + c = 3 \]
\[ 25a + 5b + c = 3 \]
Now we have the system of equations:
\[ 1. \quad a + b + c = 3 \]
\[ 2. \quad 25a + 5b + c = 3 \]
To find the vertex of the quadratic function, we know it occurs at \( x = -\frac{b}{2a} \). To find \( b \) and \( a \), we can subtract equation 1 from equation 2:
\[ (25a + 5b + c) - (a + b + c) = 3 - 3 \]
\[ 24a + 4b = 0 \]
\[ 6a + b = 0 \]
\[ b = -6a \]
Substitute \( b = -6a \) into equation 1:
\[ a - 6a + c = 3 \]
\[ -5a + c = 3 \]
\[ c = 5a + 3 \]
Now we substitute \( b = -6a \) and \( c = 5a + 3 \) into equation \( a + b + c = 3 \):
\[ a - 6a + 5a + 3 = 3 \]
\[ 0a + 3 = 3 \]
This equation holds true for any \( a \). Therefore, \( a \) can be any real number. Let's choose \( a = 1 \):
\[ b = -6 \cdot 1 = -6 \]
\[ c = 5 \cdot 1 + 3 = 8 \]
Thus, the quadratic function is \( f(x) = x^2 - 6x + 8 \).
To find the x-coordinate of the vertex:
\[ x = -\frac{b}{2a} = -\frac{-6}{2 \cdot 1} = 3 \]
Therefore, the x-coordinate of the vertex of the graph \( f \) is \( \boxed{3} \).
