The equation given is a quadratic function:

$Y = -x^2 + 6x - 8$

Key features of this quadratic function:

1. Vertex Form: To analyze the function, it's helpful to convert it into vertex form $Y = a(x-h)^2 + k$, where $(h, k)$ is the vertex of the parabola.

2. Finding the Vertex:

   • The x-coordinate of the vertex can be found using the formula $x = -\frac{b}{2a}$.
   • Here, $a = -1$ and $b = 6$. $x = -\frac{6}{2(-1)} = 3$
   • Substitute $x = 3$ back into the equation to find the y-coordinate of the vertex: $Y = -(3)^2 + 6(3) - 8 = -9 + 18 - 8 = 1$
   • So, the vertex is $(3, 1)$.

3. Axis of Symmetry: The axis of symmetry is the vertical line $x = 3$.

4. Direction of the Parabola: Since the coefficient of $x^2$ is negative, the parabola opens downward.

5. Y-intercept: When $x = 0$: $Y = -0^2 + 6(0) - 8 = -8$ So, the y-intercept is $(0, -8)$.

6. Roots/Zeroes: To find the roots, set $Y = 0$ and solve for $x$: $-x^2 + 6x - 8 = 0$ We can solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Substituting $a = -1$, $b = 6$, and $c = -8$: $x = \frac{-6 \pm \sqrt{6^2 - 4(-1)(-8)}}{2(-1)} = \frac{-6 \pm \sqrt{36 - 32}}{-2} = \frac{-6 \pm 2}{-2}$ The roots are: $x_1 = 2 \quad \text{and} \quad x_2 = 4$ So, the roots are $x = 2$ and $x = 4$.

Summary:

• Vertex: $(3, 1)$
• Axis of Symmetry: $x = 3$
• Direction: Downward (opens downward)
• Y-intercept: $(0, -8)$
• Roots: $x = 2$, $x = 4$

Would you like more details or have any questions?

Related Questions:

1. How would the graph change if the coefficient of $x^2$ was positive?
2. What is the significance of the vertex in the context of a quadratic function?
3. How do you derive the quadratic formula?
4. Can you find the focus and directrix of this parabola?
5. What are the applications of quadratic functions in real-world problems?

Tip: The vertex of a parabola gives you the maximum or minimum point on the graph, depending on the direction it opens.