To analyze the quadratic equation \( y = -6x^2 + 3x + 8 \), let's identify its key characteristics and then describe its parabolic shape.
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Determine the Vertex: The vertex of a parabola given in the form \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex: \[ x = -\frac{b}{2a} \] Here, \( a = -6 \) and \( b = 3 \).
\[ x = -\frac{3}{2 \times -6} = -\frac{3}{-12} = \frac{1}{4} \]
Now, substitute \( x = \frac{1}{4} \) back into the equation to find the y-coordinate of the vertex: \[ y = -6\left(\frac{1}{4}\right)^2 + 3\left(\frac{1}{4}\right) + 8 \] \[ y = -6\left(\frac{1}{16}\right) + \frac{3}{4} + 8 \] \[ y = -\frac{6}{16} + \frac{3}{4} + 8 \] \[ y = -\frac{3}{8} + \frac{6}{8} + \frac{64}{8} \] \[ y = \frac{67}{8} \]
So the vertex of the parabola is at \( \left( \frac{1}{4}, \frac{67}{8} \right) \).
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Direction of Opening: The parabola opens downward because the coefficient of \( x^2 \) (which is \( -6 \)) is negative.
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Y-intercept: To find the y-intercept, let \( x = 0 \): \[ y = -6(0)^2 + 3(0) + 8 = 8 \] Hence, the y-intercept is \( (0, 8) \).
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X-intercepts (if any): To find the x-intercepts, set \( y = 0 \): \[ 0 = -6x^2 + 3x + 8 \] Rearranging gives: \[ 6x^2 - 3x - 8 = 0 \] We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 6 \), \( b = -3 \), and \( c = -8 \). \[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(6)(-8)}}{2(6)} \] \[ x = \frac{3 \pm \sqrt{9 + 192}}{12} \] \[ x = \frac{3 \pm \sqrt{201}}{12} \]
The roots (x-intercepts) are: \[ x = \frac{3 + \sqrt{201}}{12} \quad \text{and} \quad x = \frac{3 - \sqrt{201}}{12} \]
- Summary: The quadratic function \( y = -6x^2 + 3x + 8 \) represents a downward-opening parabola with:
- Vertex at \( \left( \frac{1}{4}, \frac{67}{8} \right) \)
- Y-intercept at \( (0, 8) \)
- X-intercepts at \( \frac{3 \pm \sqrt{201}}{12} \)
You can sketch the parabola using this information. The vertex will be the peak of the parabola, and it will open downward due to the negative leading coefficient.