Given the function f(x) = x2 − 2x − 3, determine its range, x-intercepts (D and E), and the equation of line g passing through P and E.

From the graph, D is the negative x-intercept and E is the positive x-intercept.
The coordinates are $D(-1; 0)$ and $E(3; 0)$.

P is the y-intercept of $f(x)$. To find P, set $x=0$ in $f(x)$.

So, P is $(0; -3)$.

The line $g(x)$ passes through $P(0; -3)$ and $E(3; 0)$.
The y-intercept $c$ is the y-coordinate of P, so $c = -3$.

The graphs intersect at P$(0; -3)$ and E$(3; 0)$. These are the points where $f(x) = g(x)$.

From the graph, $f(x)$ is above $g(x)$ for $x$-values to the left of P and to the right of E.
So, $x < 0$ or $x > 3$.

The vertical distance is $D(x) = |h(x) - g(x)|$.

This is a downward-opening parabola. Its maximum value occurs at its vertex.
The x-coordinate of the vertex is $x = -\frac{b}{2a} = -\frac{1}{2(-1)} = \frac{1}{2}$.
The y-coordinate of the vertex is $k\left(\frac{1}{2}\right) = -\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right) + 6 = -\frac{1}{4} + \frac{2}{4} + \frac{24}{4} = \frac{25}{4} = 6.25$.
Since $x=\frac{1}{2}$ is within the interval $[-2; 3]$, the maximum value of $k(x)$ is 6.25.

At $x=-2$: $k(-2) = -(-2)^2 + (-2) + 6 = -4 - 2 + 6 = 0$.
At $x=3$: $k(3) = -(3)^2 + (3) + 6 = -9 + 3 + 6 = 0$.

The maximum value of $k(x)$ is 6.25, and since $k(x)$ is positive on the interval $(-2, 3)$, $D(x) = k(x)$.
The maximum vertical distance is 6.25.

This is the x-coordinate of the point of tangency.

Substitute $x = \