Find the values of $p$ for which the line $y = x + 1$ is a tangent to the curve $y = x^2 + (2p + 3)x + p + 4$. With the values of $p$ found in part (i), find the coordinates of the points where the line meets the curve.

1. Equate the equations of the line and the curve to find their points of intersection.

$x^2 + (2p + 3)x + p + 4 = x + 1$

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

$x^2 + (2p + 3 - 1)x + p + 4 - 1 = 0$
$x^2 + (2p + 2)x + p + 3 = 0$

3. Apply the condition for a line to be a tangent to a curve (discriminant equals zero).

$b^2 - 4ac = 0$
$(2p + 2)^2 - 4(1)(p + 3) = 0$

4. Solve the resulting quadratic equation for $p$.

$4p^2 + 8p + 4 - 4p - 12 = 0$
$4p^2 + 4p - 8 = 0$
$p^2 + p - 2 = 0$
$(p + 2)(p - 1) = 0$
$p = -2 \text{ or } p = 1$

1. State the values of $p$ from part (i) and the intersection equation

From part (i), the values of $p$ are $p = -2$ and $p = 1$.
The intersection equation is $x^2 + (2p + 2)x + p + 3 = 0$.

2. Find the coordinates for $p = -2$

Substitute $p = -2$ into the intersection equation:
$x^2 + (2(-2) + 2)x + (-2) + 3 = 0$
$x^2 - 2x + 1 = 0$
$(x - 1)^2 = 0$
$x = 1$
Substitute $x = 1$ into the line equation $y = x + 1$:
$y = 1 + 1 = 2$
The point is $(1, 2)$.

3. Find the coordinates for $p = 1$

Substitute $p = 1$ into the intersection equation:
$x^2 + (2(1) + 2)x + 1 + 3 = 0$
$x^2 + 4x + 4 = 0$
$(x + 2)^2 = 0$
$x = -2$
Substitute $x = -2$ into the line equation $y = x + 1$:
$y = -2 + 1 = -1$
The point is $(-2, -1)$.