Find the equations of the following functions:

The formula for the slope is $m = \frac{y_2 - y_1}{x_2 - x_1}$.
Let $(x_1, y_1) = (-2, -1)$ and $(x_2, y_2) = (1, 4)$.

$$m = \frac{4 - (-1)}{1 - (-2)} = \frac{4 + 1}{1 + 2} = \frac{5}{3}$$

We will use the slope-intercept form.
Substitute the slope $m = \frac{5}{3}$ and one of the points, for example $(1, 4)$, into $y = mx + c$.

$$4 = \frac{5}{3}(1) + c$$ $$4 = \frac{5}{3} + c$$ $$c = 4 - \frac{5}{3}$$ $$c = \frac{12}{3} - \frac{5}{3}$$ $$c = \frac{7}{3}$$

Substitute $m = \frac{5}{3}$ and $c = \frac{7}{3}$ into $y = mx + c$.

$$y = \frac{5}{3}x + \frac{7}{3}$$

The equation of the line is $\boxed{y = \frac{5}{3}x + \frac{7}{3}}$.

From the graph, the vertex is $(h, k) = (4, -1)$.

$$y = a(x - 4)^2 - 1$$

From the graph, the parabola appears to pass through the point $(5, 0)$ on the x-axis.
Substitute $(x, y) = (5, 0)$ into the equation.

$$0 = a(5 - 4)^2 - 1$$ $$0 = a(1)^2 - 1$$ $$0 = a - 1$$ $$a = 1$$

Substitute $a = 1$ into the vertex form.

$$y = 1(x - 4)^2 - 1$$ $$y = (x - 4)^2 - 1$$

The equation of the parabola is $\boxed{y = (x - 4)^2 - 1}$.