Given y = 2x^2 – 4x-1 , find the vertex, focus and the equation of the directrix line.

First, we need to write the equation of the parabola in standard form: {eq}y = a(x - h)^2 + k {/eq}, where (h, k) is the vertex. Completing the square, we have: {eq}y = 2(x^2 - 2x) - 1 {/eq} {eq}y = 2(x^2 - 2x + 1) - 2 + 1 {/eq} {eq}y = 2(x - 1)^2 - 1 {/eq} So the vertex is at (1, -1). Next, we can use the formula {eq}4p = a {/eq} to find the distance from the vertex to the focus and the equation of the directrix line. Since {eq}a = 2 {/eq}, we have {eq}4p = 2 {/eq}, so {eq}p = \frac{1}{2} {/eq}. The focus is {eq}(h, k + p) = (1, -1 + \frac{1}{2}) = (1, -frac{1}{2}) {/eq}. The directrix line is a horizontal line that is {eq}p {/eq} units below the vertex, so it is {eq}y = k - p = -1 - \frac{1}{2} = -frac{3}{2} {/eq}. Therefore, the vertex is (1, -1), the focus is (1, -1/2), and the equation of the directrix line is {eq}y = -frac{3}{2} {/eq}.