Find the vertex of the following quadratic function

y = 2x^{2}
– 4x^{ }+ 5

Vertex: (^{},
^{})

In order to find the vertex, we first find the axis of symmetry

x = = = 1

Now, we know that the x coordinate of the vertex is 1, so we can plug this in and solve for the y coordinate

y = 2(1)^{2}
– 4(1) + 5 = 2 – 4 + 5 = 3

So the vertex is at (1, 3)