To find the vertex of the quadratic equation \( y = 4x^2 + 3x - 8 \), we can use the vertex formula for a parabola given by the equation \( y = ax^2 + bx + c \). The x-coordinate of the vertex is calculated using the formula:
\[ x = -\frac{b}{2a} \]
In this case:
- \( a = 4 \)
- \( b = 3 \)
- \( c = -8 \)
Substituting the values of \( a \) and \( b \):
\[ x = -\frac{3}{2 \cdot 4} = -\frac{3}{8} \]
Next, we need to calculate the y-coordinate by substituting \( x = -\frac{3}{8} \) back into the original equation:
\[ y = 4\left(-\frac{3}{8}\right)^2 + 3\left(-\frac{3}{8}\right) - 8 \] \[ = 4 \cdot \frac{9}{64} - \frac{9}{8} - 8 \] \[ = \frac{36}{64} - \frac{72}{64} - \frac{512}{64} \] \[ = \frac{36 - 72 - 512}{64} \] \[ = \frac{-548}{64} = -8.5625 \]
Thus, the vertex of the quadratic \( y = 4x^2 + 3x - 8 \) is approximately:
\[ \left(-0.375, -8.563\right) \]
So, the correct response from the given options is:
\(\left(-0.375, -8.563\right)\)