Question

find the y - intercept, the axis of symmetry, and the vertex of the graph of the function.\\(f(x)=x^{2}-6x + 11\\)\\(\\)\\(\\)the y - intercept is \\((0,11)\\). (type an ordered pair.)\\(\\)the axis of symmetry is \\(x = 3\\). (simplify your answer. type an equation.)\\(\\)the vertex is \\(\square\\). (type an ordered pair.)

Explanation:

Step1: Recall vertex formula for parabola

For a quadratic function \( f(x) = ax^2 + bx + c \), the x - coordinate of the vertex is given by \( x = -\frac{b}{2a} \). We already know from the axis of symmetry that \( x = 3 \) (since the axis of symmetry of a parabola \( y=ax^{2}+bx + c \) is \( x=-\frac{b}{2a} \), and for \( f(x)=x^{2}-6x + 11 \), \( a = 1 \), \( b=-6 \), so \( x=-\frac{-6}{2\times1}=3 \)).

Step2: Find the y - coordinate of the vertex

To find the y - coordinate of the vertex, we substitute \( x = 3 \) into the function \( f(x)=x^{2}-6x + 11 \).
Substitute \( x = 3 \) into \( f(x) \):
\( f(3)=(3)^{2}-6\times(3)+11 \)
First, calculate \( (3)^{2}=9 \), \( 6\times(3) = 18 \).
Then \( f(3)=9-18 + 11 \)
\( 9-18=-9 \), and \( -9 + 11 = 2 \).

Answer:

\((3,2)\)