The given function is:

$r(x) = -x^2 + 6x - 8$

We need to find two points on the graph of this parabola other than the vertex and the $x$-intercepts. From the image, the vertex is $(3, 1)$ and the $x$-intercepts are $(4, 0)$ and $(2, 0)$.

To find two other points, we can choose random values for $x$, substitute them into the equation for $r(x)$, and calculate the corresponding $y$-values.

Let’s choose $x = 1$ and $x = 5$.

Step 1: For $x = 1$

$r(1) = -(1)^2 + 6(1) - 8 = -1 + 6 - 8 = -3$ So, one point is $(1, -3)$.

Step 2: For $x = 5$

$r(5) = -(5)^2 + 6(5) - 8 = -25 + 30 - 8 = -3$ So, another point is $(5, -3)$.

Thus, the two points on the graph of the parabola are $(1, -3)$ and $(5, -3)$.

Would you like more details or have any questions?

Here are some related questions:

1. What are the steps to find the vertex of a quadratic function?
2. How can we determine the $x$-intercepts of a quadratic function?
3. Why do parabolas exhibit symmetry around the vertex?
4. How do you determine whether a parabola opens upward or downward?
5. What is the importance of the discriminant in finding the roots of a quadratic function?

Tip: The vertex form of a quadratic function makes it easier to find the vertex and understand the graph's shape.