y=-3x²-6x+2
The equation you've provided is a quadratic function in the form \( y = ax^2 + bx + c \), where:
- \( a = -3 \)
- \( b = -6 \)
- \( c = 2 \)
Characteristics of the Quadratic Function
Direction of the Parabola: Since \( a < 0 \), the parabola opens downwards.
Vertex: The vertex of the parabola can be found using the formula \( x = -\frac{b}{2a} \).
Substituting the values of \( a \) and \( b \): \[ x = -\frac{-6}{2 \times -3} = \frac{6}{-6} = -1 \]
To find the corresponding \( y \)-coordinate, substitute \( x = -1 \) back into the equation: \[ y = -3(-1)^2 - 6(-1) + 2 = -3(1) + 6 + 2 = -3 + 6 + 2 = 5 \]
So the vertex is at \( (-1, 5) \).
Y-intercept: The y-intercept occurs when \( x = 0 \): \[ y = -3(0)^2 - 6(0) + 2 = 2 \] Thus, the y-intercept is at the point \( (0, 2) \).
X-intercepts: To find the x-intercepts, set \( y = 0 \): \[ 0 = -3x^2 - 6x + 2 \] This can be solved using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(-3)(2)}}{2(-3)} = \frac{6 \pm \sqrt{36 + 24}}{-6} = \frac{6 \pm \sqrt{60}}{-6} = \frac{6 \pm 2\sqrt{15}}{-6} \] This simplifies to: \[ x = -1 \mp \frac{\sqrt{15}}{3} \] So, the x-intercepts are \( x = -1 + \frac{\sqrt{15}}{3} \) and \( x = -1 - \frac{\sqrt{15}}{3} \).
Summary of Key Points:
- Vertex: \( (-1, 5) \)
- Y-intercept: \( (0, 2) \)
- X-intercepts: \( \left(-1 + \frac{\sqrt{15}}{3}, 0\right) \) and \( \left(-1 - \frac{\sqrt{15}}{3}, 0\right) \)
- The parabola opens downwards.
If you'd like further analysis or to see a graph of this function, just let me know!
