Answer
The x-value of the vertex was incorrectly evaluated, and should be x = +1 instead, as $x = \frac{-b}{2a}$
The proper vertex is (1,-5), so this graph should be transformed two units to the right and 5 units down.
Work Step by Step
In this standard-form parabola of the form $ax^2+bx+c$, a = 2, b = -4, and c = -3.
The x-value of the vertex is $x = \frac{-b}{2a}$, but here, they seemed to evaluate $\frac{+b}{2a}$.
The true x-value is: $x = \frac{-b}{2a} = \frac{-(-4)}{2*2} = 1$
Plugging in, the y-value is: $y = 2x^2 - 4x -3 = 2(1)^2-4(1) -3 = -5
Vertex is (1,-5)
