To find the vertex of the quadratic function \(y = 4x^2 + 3x - 8\), we can use the vertex formula for a parabola given in standard form \(y = ax^2 + bx + c\). The x-coordinate of the vertex can be found using the formula:
\[ x = -\frac{b}{2a} \]
Here, \(a = 4\) and \(b = 3\). Plugging these values into the formula:
\[ x = -\frac{3}{2 \cdot 4} = -\frac{3}{8} \]
Next, we need to find the y-coordinate of the vertex by substituting \(x = -\frac{3}{8}\) back into the original quadratic equation:
\[ y = 4\left(-\frac{3}{8}\right)^2 + 3\left(-\frac{3}{8}\right) - 8 \]
Calculating \( \left(-\frac{3}{8}\right)^2 \):
\[ \left(-\frac{3}{8}\right)^2 = \frac{9}{64} \]
Now substituting this value:
\[ y = 4 \cdot \frac{9}{64} + 3 \cdot \left(-\frac{3}{8}\right) - 8 \]
\[ y = \frac{36}{64} - \frac{9}{8} - 8 \]
Now converting everything to a common denominator of 64:
\[ -\frac{9}{8} = -\frac{72}{64}, \quad -8 = -\frac{512}{64} \]
Now substituting back in:
\[ y = \frac{36}{64} - \frac{72}{64} - \frac{512}{64} = \frac{36 - 72 - 512}{64} = \frac{-548}{64} = -\frac{137}{16} \]
Thus, the vertex of the quadratic \(y = 4x^2 + 3x - 8\) is at:
\[ \left(-\frac{3}{8}, -\frac{137}{16}\right) \]
Among the given choices, the only one that indicates this point is:
\[ (-0.375, -8.563) \]
which approximately matches the calculated vertex coordinates, as \(-0.375\) is \(-\frac{3}{8}\) and \(-8.563\) is close to \(-\frac{137}{16}\).
Therefore, the answer is:
(−0.375,−8.563)