the axis of symmetry for the function $f(x)=-2x^{2}+4x + 1$ is the line $x = 1$. where is the vertex of the…

the axis of symmetry for the function $f(x)=-2x^{2}+4x + 1$ is the line $x = 1$. where is the vertex of the function located?\n(0, 1)\n(1, 3)\n(1, 7)\n(2, 1)

the axis of symmetry for the function $f(x)=-2x^{2}+4x + 1$ is the line $x = 1$. where is the vertex of the function located?\n(0, 1)\n(1, 3)\n(1, 7)\n(2, 1)

Answer

Explanation:

Step1: Recall vertex - axis of symmetry relation

The x - coordinate of the vertex of a parabola (y = ax^{2}+bx + c) is on the axis of symmetry. Given axis of symmetry (x = 1), so (x) - coordinate of vertex is (x=1).

Step2: Find y - coordinate of vertex

Substitute (x = 1) into (f(x)=-2x^{2}+4x + 1). Then (f(1)=-2(1)^{2}+4(1)+1). [ \begin{align*} f(1)&=-2\times1 + 4\times1+1\ &=-2 + 4+1\ &=3 \end{align*} ]

Answer:

B. ((1,3))