To find two points on the graph of the parabola defined by the function $k(x) = x^2 + 5x + 4$ other than the vertex and $x$-intercepts, we can choose any values of $x$ that are different from the $x$-intercepts, $x = -4$ and $x = -1$.

Let's choose $x = 0$ and $x = 1$ as our points.

1. For $x = 0$:

   Substitute $x = 0$ into the function:

   $k(0) = (0)^2 + 5(0) + 4 = 0 + 0 + 4 = 4$

   So, the point is $(0, 4)$.

2. For $x = 1$:

   Substitute $x = 1$ into the function:

   $k(1) = (1)^2 + 5(1) + 4 = 1 + 5 + 4 = 10$

   So, the point is $(1, 10)$.

Summary:

Two points on the graph of the parabola other than the vertex and $x$-intercepts are $(0, 4)$ and $(1, 10)$.

Would you like more details or have any questions?

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Here are some related questions you might consider:

1. How do you find the vertex of the parabola $k(x) = x^2 + 5x + 4$?
2. What is the axis of symmetry for this parabola?
3. How can you determine the direction in which the parabola opens?
4. What is the y-intercept of the function $k(x) = x^2 + 5x + 4$?
5. How do you graph the entire function $k(x) = x^2 + 5x + 4$?
6. How do you find the minimum or maximum value of $k(x)$?
7. How do the points $(0, 4)$ and $(1, 10)$ relate to the shape of the parabola?
8. Can you express the function $k(x)$ in factored form?

Tip: Choosing points near the vertex or intercepts helps in sketching a more accurate graph of the parabola.