To find two points on the graph of the parabola represented by the function $k(x) = -x^2 + 6x - 8$ other than the vertex and $x$-intercepts, we can choose any $x$-values and compute the corresponding $y$-values using the function $k(x)$.

Step-by-Step Solution:

1. Choose two $x$-values:

   • We can choose any two $x$-values that are not the $x$-intercepts or the $x$-coordinate of the vertex. To avoid those, let's first find the vertex and the $x$-intercepts.

2. Find the vertex: The vertex of a parabola given by $ax^2 + bx + c$ occurs at $x = -\frac{b}{2a}$.

   For $k(x) = -x^2 + 6x - 8$:

   • $a = -1$
   • $b = 6$

   So, the $x$-coordinate of the vertex is: $x = -\frac{6}{2(-1)} = \frac{6}{2} = 3$

   The $x$-coordinate of the vertex is $x = 3$.

3. Find the $x$-intercepts: The $x$-intercepts occur where $k(x) = 0$. To find them, solve the equation: $-x^2 + 6x - 8 = 0$

   Rearrange and factor the quadratic: $x^2 - 6x + 8 = 0$ $(x - 2)(x - 4) = 0$

   So, the $x$-intercepts are at $x = 2$ and $x = 4$.

4. Choose two other $x$-values: We need to choose two $x$-values different from 2, 3, and 4. Let's choose $x = 0$ and $x = 1$.

5. Compute the corresponding $y$-values:

   • For $x = 0$: $k(0) = -0^2 + 6(0) - 8 = -8$ So, one point on the graph is $(0, -8)$.

   • For $x = 1$: $k(1) = -(1)^2 + 6(1) - 8 = -1 + 6 - 8 = -3$ So, another point on the graph is $(1, -3)$.

Conclusion:

The two points on the graph of the parabola $k(x) = -x^2 + 6x - 8$, other than the vertex and the $x$-intercepts, are:

• $(0, -8)$
• $(1, -3)$

Would you like further clarification, or do you have any additional questions?

Related Questions:

1. How do you find the $y$-intercept of a parabola?
2. What is the axis of symmetry for the parabola given by $k(x) = -x^2 + 6x - 8$?
3. How do you determine whether a parabola opens upwards or downwards?
4. How would you complete the square to find the vertex form of $k(x)$?
5. What is the significance of the discriminant in the context of finding the $x$-intercepts?

Tip:

When working with parabolas, knowing the vertex and the direction it opens helps in quickly sketching its graph. The $x$-intercepts provide additional points of intersection with the x-axis, making it easier to visualize.