Find the vertex and axis of symmetry of the graph of the function.
-$$f ( x ) = 2 x ^ { 2 } - 4 x - 1$$
A) (-1, 5) ; x = -1
B) (-2, 15) ; x = -2
C) (1, -3) ; x = 1
D) (2, 3) ; x = 2

Question

Quizplus · Accepted Answer

The vertex of a parabola in the form $f(x) = ax^2 + bx + c$ can be found using the formula $x = -\frac{b}{2a}$. For $f(x) = 2x^2 - 4x - 1$, $a = 2$ and $b = -4$, so $x = -\frac{-4}{2*2} = 1$. Plugging $x = 1$ back into the function gives $f(1) = 2(1)^2 - 4(1) - 1 = -3$. Therefore, the vertex is $(1, -3)$, and the axis of symmetry is $x = 1$.

Find the Vertex and Axis of Symmetry of the Graph f(x)=2x2−4x−1f ( x ) = 2 x ^ { 2 } - 4 x - 1f(x)=2x2−4x−1

Find the Vertex and Axis of Symmetry of the Graph $f ( x ) = 2 x ^ { 2 } - 4 x - 1$