Find the range of values of $k$ for which the line $2x - y = 5$ intersects the curve $xy = kx - 2$ at two distinct points. Find the smallest integer value of $h$ for which the graph $y = 2x^2 - 4x + h$ lies entirely above the line $y = 3$ for all values of $x$.

1. Express $y$ in terms of $x$ from the linear equation and substitute into the curve's equation

From $2x - y = 5$, we have $y = 2x - 5$.
Substitute this into $xy = kx - 2$:
$x(2x - 5) = kx - 2$

2. Rearrange the equation into standard quadratic form $ax^2 + bx + c = 0$

$2x^2 - 5x = kx - 2$
$2x^2 - 5x - kx + 2 = 0$
$2x^2 - (5 + k)x + 2 = 0$

3. Apply the discriminant condition for two distinct points of intersection

For two distinct points of intersection, the discriminant $b^2 - 4ac > 0$.
$(-(5 + k))^2 - 4(2)(2) > 0$
$(5 + k)^2 - 16 > 0$

4. Solve the quadratic inequality to find the range of $k$

$k^2 + 10k + 25 - 16 > 0$
$k^2 + 10k + 9 > 0$
$(k + 9)(k + 1) > 0$
Therefore, $k < -9$ or $k > -1$.

1. Set up the inequality for the graph to lie entirely above the line

For the graph $y = 2x^2 - 4x + h$ to lie entirely above $y = 3$, we must have:
$2x^2 - 4x + h > 3$
$2x^2 - 4x + (h - 3) > 0$ for all real values of $x$.

2. Apply the discriminant condition for a quadratic to be always positive

Since the coefficient of $x^2$ is $2 > 0$, the curve opens upwards. For the quadratic expression to be always positive, it must have no real roots, so its discriminant must be less than zero:
$b^2 - 4ac < 0$
$(-4)^2 - 4(2)(h - 3) < 0$
$16 - 8(h - 3) < 0$
$16 - 8h + 24 < 0$
$40 - 8h < 0$

3. Solve the inequality for $h$ and find the smallest integer

$8h > 40$
$h > 5$

The smallest integer value of $h$ that satisfies this inequality is $6$.