To find the vertex of the quadratic function [tex]\( y = x^2 - 6x + 1 \)[/tex], we can use the vertex formula for a parabola. The vertex of a parabola described by the quadratic equation [tex]\( y = ax^2 + bx + c \)[/tex] can be found using the following steps:
1. Identify the coefficients: From the given quadratic equation [tex]\( y = x^2 - 6x + 1 \)[/tex], we have:
- [tex]\( a = 1 \)[/tex] (coefficient of [tex]\( x^2 \)[/tex])
- [tex]\( b = -6 \)[/tex] (coefficient of [tex]\( x \)[/tex])
- [tex]\( c = 1 \)[/tex] (constant term)
2. Calculate the x-coordinate of the vertex: The x-coordinate of the vertex can be found using the formula:
[tex]\[
x = -\frac{b}{2a}
\][/tex]
Substituting the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] into the formula, we get:
[tex]\[
x = -\frac{-6}{2 \cdot 1} = \frac{6}{2} = 3
\][/tex]
3. Calculate the y-coordinate of the vertex: To find the y-coordinate, we substitute the x-coordinate back into the original quadratic equation [tex]\( y = x^2 - 6x + 1 \)[/tex]:
[tex]\[
y = (3)^2 - 6(3) + 1 = 9 - 18 + 1 = -8
\][/tex]
Therefore, the vertex of the quadratic function [tex]\( y = x^2 - 6x + 1 \)[/tex] is at the point [tex]\( (3, -8) \)[/tex].
The correct answer is:
D. [tex]\( (3, -8) \)[/tex]
