To analyze the quadratic equation \( y = -6x^2 + 3x + 8 \) and determine the vertex, we first note that this is a downward-facing parabola since the coefficient of \( x^2 \) (which is -6) is negative. The vertex of a parabola in the form \( y = ax^2 + bx + c \) is given by the formula:
\[ x = -\frac{b}{2a} \]
Here, \( a = -6 \) and \( b = 3 \).
Substituting in:
\[ x = -\frac{3}{2 \times -6} = -\frac{3}{-12} = \frac{1}{4} \]
Next, to find the corresponding \( y \)-value of the vertex, we substitute \( x = \frac{1}{4} \) back into the equation:
\[ y = -6\left(\frac{1}{4}\right)^2 + 3\left(\frac{1}{4}\right) + 8 \]
Calculating this step-by-step:
\[ y = -6\left(\frac{1}{16}\right) + \frac{3}{4} + 8 \]
\[ y = -\frac{6}{16} + \frac{3}{4} + 8 \]
\[ y = -\frac{3}{8} + \frac{6}{8} + 8 \]
\[ y = \frac{3}{8} + 8 = \frac{3}{8} + \frac{64}{8} = \frac{67}{8} = 8.375 \]
So, the vertex (the low point) of the parabola is:
\[ \left(\frac{1}{4}, 8.375\right) \]
Now, since this is a downward-facing parabola, it has a maximum point at the vertex, not a minimum. Therefore, we can summarize:
- The low point on the parabola is \((\frac{1}{4}, 8.375)\).
- The high point does not exist in this case, as the vertex represents the maximum point.
In conclusion, based on your statements, here are the final values:
- The vertex is \((\frac{1}{4}, 8.375)\), and this is the high point for the equation \( y = -6x^2 + 3x + 8 \). The values \( (-1, -1) \) and \( (0.5, 9) \) listed in your question seem to be unrelated to this function either mathematically or contextually since they do not match the vertex or corresponding calculations.
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