Identify the vertex, axis of symmetry, and intercepts for the graph of the function.
-$$g ( x ) = - x ^ { 2 } + 6 x - 11$$
A) Vertex at $( - 3 , - 38 )$; axis: $x = - 3 ; x$-intercepts: $( 3 - \sqrt { 2 } , 0 )$ and $( 3 + \sqrt { 2 } , 0 ) ; y$-intercept: $( 0 , - 11 )$
B) Vertex at $( - 3 , - 38 )$; axis: $y = - 38 ; x$-intercepts: none; $y$-intercept: $( 0,11 )$
C) Vertex at $( 3 , - 2 )$; axis: $y = - 2 ; x$-intercepts: $( 3 - \sqrt { 2 } , 0 )$ and $( 3 + \sqrt { 2 } , 0 ) ; y$-intercept: $( 0,11 )$
D) Vertex at $( 3 , - 2 )$; axis: $x = 3 ; x$-intercepts: none; $y$-intercept: $( 0 , - 11 )$

Question

Quizplus · Accepted Answer

The vertex of a parabola in the form $y = ax^2 + bx + c$ can be found using the formula $(-b/2a, f(-b/2a))$. For $g(x) = -x^2 + 6x - 11$, the vertex is at $(3, -2)$. The axis of symmetry is $x = 3$. There are no real x-intercepts because the discriminant $b^2 - 4ac$ is negative, indicating no real roots. The y-intercept is found by evaluating $g(0)$, which gives $(0, -11)$.

Identify the Vertex, Axis of Symmetry, and Intercepts for the Graph