To find the vertex of the quadratic function \( f(x) = (x - 8)(x - 2) \), we can first rewrite it in standard form \( ax^2 + bx + c \).
-
Expand the expression: \[ f(x) = (x - 8)(x - 2) = x^2 - 2x - 8x + 16 = x^2 - 10x + 16 \]
-
Identify the coefficients: From \( f(x) = x^2 - 10x + 16 \),
- \( a = 1 \)
- \( b = -10 \)
- \( c = 16 \)
-
Use the vertex formula: The x-coordinate of the vertex of a quadratic function in the form \( ax^2 + bx + c \) is given by: \[ x = -\frac{b}{2a} \] Plugging in our values: \[ x = -\frac{-10}{2 \cdot 1} = \frac{10}{2} = 5 \]
-
Find the y-coordinate: To find the y-coordinate of the vertex, substitute \( x = 5 \) back into the function: \[ f(5) = (5 - 8)(5 - 2) = (-3)(3) = -9 \]
-
Vertex: Therefore, the vertex of the quadratic function \( f(x) = (x - 8)(x - 2) \) is: \[ (5, -9) \]