Find the vertex and axis of symmetry of the graph of the function.
-$$f ( x ) = - 2 x ^ { 2 } + 4 x + 1$$
A)vertex: (-1, -5); x = -1
B)vertex: (1, 3); x = 1
C)vertex: (1, 7); x = 1
D)vertex: (-1, 3); x = -1

Question

Quizplus · Accepted Answer

The vertex of a parabola given by $f(x) = ax^2 + bx + c$ can be found using the formula $x = -\frac{b}{2a}$. For the given function, $a = -2$ and $b = 4$, so the x-coordinate of the vertex is $x = -\frac{4}{2(-2)} = 1$. Plugging this back into the function gives the y-coordinate: $f(1) = -2(1)^2 + 4(1) + 1 = 3$. Therefore, the vertex is at (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 $f ( x ) = - 2 x ^ { 2 } + 4 x + 1$