Answer :
Answer:
To determine which graph represents the quadratic function \(-x^2 + 6x - 1\), let's analyze its features:
1. **Direction**: Since the coefficient of \(x^2\) is negative (\(-1\)), the parabola opens downward.
2. **Vertex**: The vertex form of a quadratic equation \(ax^2 + bx + c\) can be found using the vertex formula \(x = -\frac{b}{2a}\). For \(-x^2 + 6x - 1\):
\[
x = -\frac{6}{2 \cdot (-1)} = \frac{6}{2} = 3
\]
Plug \(x = 3\) back into the equation to find the y-coordinate of the vertex:
\[
y = -(3)^2 + 6(3) - 1 = -9 + 18 - 1 = 8
\]
So, the vertex is at \((3, 8)\).
3. **Y-intercept**: The y-intercept is the value of the function when \(x = 0\):
\[
y = -0^2 + 6(0) - 1 = -1
\]
So, the y-intercept is \((0, -1)\).
4. **X-intercepts**: To find the x-intercepts (where the function crosses the x-axis), set the equation equal to zero and solve for \(x\):
\[
-x^2 + 6x - 1 = 0
\]
Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[
a = -1, \quad b = 6, \quad c = -1
\]
\[
x = \frac{-6 \pm \sqrt{6^2 - 4(-1)(-1)}}{2(-1)} = \frac{-6 \pm \sqrt{36 - 4}}{-2} = \frac{-6 \pm \sqrt{32}}{-2} = \frac{-6 \pm 4\sqrt{2}}{-2}
\]
\[
x = \frac{-6 + 4\sqrt{2}}{-2} \text{ and } x = \frac{-6 - 4\sqrt{2}}{-2}
\]
Simplifying:
\[
x = 3 - 2\sqrt{2} \text{ and } x = 3 + 2\sqrt{2}
\]
Given these characteristics, look for a graph with:
- A downward-opening parabola.
- A vertex at \((3, 8)\).
- Y-intercept at \((0, -1)\).
- X-intercepts at approximately \(3 - 2\sqrt{2}\) and \(3 + 2\sqrt{2}\).
You can graph this equation using a graphing calculator or software like Desmos to visually confirm these features.
Step-by-step explanation:
