To solve this problem, let's break down the task step-by-step.

Step 1: Analyze the Equation

The given equation is: $y = -x^2 + 6x - 5$ This equation represents a parabola that opens downward (since the coefficient of $x^2$ is negative).

Step 2: Find the Roots

To find the roots (x-intercepts) of the equation, we set $y = 0$: $-x^2 + 6x - 5 = 0$ This can be rewritten as: $x^2 - 6x + 5 = 0$ We solve this quadratic equation using the quadratic formula: $x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$ Simplifying, we get: $x = \frac{6 \pm \sqrt{36 - 20}}{2}$ $x = \frac{6 \pm \sqrt{16}}{2}$ $x = \frac{6 \pm 4}{2}$ Thus, the roots are: $x = \frac{6 + 4}{2} = 5 \quad \text{and} \quad x = \frac{6 - 4}{2} = 1$ So, the roots are $x = 5$ and $x = 1$.

Step 3: Find the Vertex

The vertex of a parabola in the form $y = ax^2 + bx + c$ can be found using the formula: $x = -\frac{b}{2a}$ Here, $a = -1$ and $b = 6$, so: $x = -\frac{6}{2 \cdot -1} = 3$ To find the y-coordinate of the vertex, substitute $x = 3$ back into the equation: $y = -(3)^2 + 6 \cdot 3 - 5$ $y = -9 + 18 - 5$ $y = 4$ So, the vertex is at $(3, 4)$.

Step 4: Plot the Points

You need to plot five points, including the roots and the vertex. The key points are:

1. Root at $(1, 0)$
2. Root at $(5, 0)$
3. Vertex at $(3, 4)$

Step 5: Choose Two Additional Points

Let's select additional points to provide a more complete graph. For example:

• When $x = 0$: $y = -(0)^2 + 6 \cdot 0 - 5 = -5$ So, $(0, -5)$ is one point.

• When $x = 6$: $y = -(6)^2 + 6 \cdot 6 - 5$ $y = -36 + 36 - 5 = -5$ So, $(6, -5)$ is another point.

Summary of Points to Plot

The points you should plot on the graph are:

1. $(1, 0)$
2. $(5, 0)$
3. $(3, 4)$ (vertex)
4. $(0, -5)$
5. $(6, -5)$

These points will outline the parabola for $y = -x^2 + 6x - 5$.

Would you like further clarification on any of these steps or details on how to complete the graph?

Here are 5 related questions for further exploration:

1. How do you determine if a parabola opens upwards or downwards?
2. What does the vertex represent in a parabolic graph?
3. How can you find the axis of symmetry for a parabola?
4. How do you use the quadratic formula to find the roots of any quadratic equation?
5. How would the graph change if the equation were $y = x^2 - 6x + 5$ instead?

Tip: When graphing parabolas, always identify the vertex and the direction of opening first—it helps in sketching the graph accurately.