- The line of symmetry for the parabola \(y = -3x^2 + 6x - 1\) is \(x = 1\).
- The line \(y = 1\) is a horizontal line crossing the y-axis at 1.
- Solving \(-3x^2 + 6x - 1 = 1\) results in the roots \(x = 1 - \frac{\sqrt{3}}{3}\) and \(x = 1 + \frac{\sqrt{3}}{3}\).
- Final Answer: \(\boxed{x = 1 - \frac{\sqrt{3}}{3}, x = 1 + \frac{\sqrt{3}}{3}}\)
Explanation
- Analysis of the Quadratic Function and Problem Requirements
We are given the quadratic function \(y = -3x^2 + 6x - 1\). From the provided data, we know its vertex is at \((1, 2)\). Our tasks are to determine the line of symmetry, identify the horizontal line \(y = 1\), and solve the equation \(-3x^2 + 6x - 1 = 1\) by finding the points where the parabola intersects the line \(y = 1\).
- Identifying the Line of Symmetry
A parabola of the form \(y = ax^2 + bx + c\) is symmetric about a vertical line passing through its vertex. Given the vertex is \((1, 2)\), the x-coordinate of the vertex is \(h = 1\). Therefore, the line of symmetry is the vertical line defined by the equation \(x = 1\). This line divides the parabola into two mirror-image halves.
- Representing the Line y = 1
The line \(y = 1\) is a horizontal line that intersects all points where the y-coordinate is exactly 1. On a coordinate plane, this is a line running parallel to the x-axis, crossing the y-axis at the point \((0, 1)\). Any intersection points between the parabola and this line represent solutions to the equation \(-3x^2 + 6x - 1 = 1\).
- Solving the Equation -3x^2 + 6x - 1 = 1
To solve the equation \(-3x^2 + 6x - 1 = 1\), we set the two expressions equal. Rearranging the terms to form a standard quadratic equation:
\[-3x^2 + 6x - 1 - 1 = 0\]
\[-3x^2 + 6x - 2 = 0\]
Multiplying by -1 for convenience, we have \(3x^2 - 6x + 2 = 0\). Applying the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a=3, b=-6, c=2\):
\[x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(2)}}{2(3)}\]
\[x = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm \sqrt{12}}{6}\]
Simplifying \(\sqrt{12} = 2\sqrt{3}\):
\[x = \frac{6 \pm 2\sqrt{3}}{6} = 1 \pm \frac{\sqrt{3}}{3}\]
Thus, the solutions are \(x = 1 - \frac{\sqrt{3}}{3}\) and \(x = 1 + \frac{\sqrt{3}}{3}\).
Examples
Quadratic functions are widely used in physics to model the trajectory of a projectile. If a ball is thrown into the air, its height over time follows a parabolic path. Finding the line of symmetry helps identify the time at which the ball reaches its maximum height (the vertex), and solving for a specific height (like the line \(y=1\)) tells you exactly when the ball reaches that height on its way up and on its way down.