How do you find the vertex and axis of symmetry, and then graph the parabola
given by: #f(x)=2x^2-4x+1#?

Question

Ethan Adkins · Accepted Answer

Graph #f(x) = 2x^2 - 4x + 1# x-coordinate of axis of symmetry and vertex: x = ( -b)/2a = 4/4 = 1 y-coordinate of vertex: f(1 ) = 2 - 4 + 1 = - 1 graph{2x^2 - 4x + 1 [-10, 10, -5, 5]}

Eva Adamson · Answer

undefined To find the vertex of the parabola represented by the function $ f(x) = 2x^2 - 4x + 1 $, use the formula: 
$$ x = \frac{-b}{2a} $$

To find the axis of symmetry, use the x-coordinate of the vertex.

Substitute this value into the original function to find the y-coordinate of the vertex.

The vertex is $(x, y)$.

The axis of symmetry is the vertical line $x = x$ where $x$ is the x-coordinate of the vertex.

To graph the parabola, plot the vertex, the axis of symmetry, and at least two additional points symmetrically placed on either side of the axis of symmetry.

Connect the points smoothly to form the parabola.

How do you find the vertex and axis of symmetry, and then graph the parabola given by: #f(x)=2x^2-4x+1#?