To find the vertex of the parabola given by the equation \( y = 14x^2 + 3x - 1 \), we can use the formula for the x-coordinate of the vertex for a quadratic equation in standard form \( y = ax^2 + bx + c \), which is given by:
\[ x = -\frac{b}{2a} \]
In this case, \( a = 14 \) and \( b = 3 \). Plugging in these values:
\[ x = -\frac{3}{2 \cdot 14} = -\frac{3}{28} \]
Next, we need to calculate the corresponding y-coordinate by substituting \( x = -\frac{3}{28} \) back into the original equation:
\[ y = 14\left(-\frac{3}{28}\right)^2 + 3\left(-\frac{3}{28}\right) - 1 \]
Calculating \( \left(-\frac{3}{28}\right)^2 \):
\[ \left(-\frac{3}{28}\right)^2 = \frac{9}{784} \]
Now substitute this back:
\[ y = 14 \cdot \frac{9}{784} + 3 \cdot \left(-\frac{3}{28}\right) - 1 \]
Calculating each term:
- \( 14 \cdot \frac{9}{784} = \frac{126}{784} = \frac{63}{392} \)
- \( 3 \cdot \left(-\frac{3}{28}\right) = -\frac{9}{28} = -\frac{126}{392} \)
Now substituting these into the equation:
\[ y = \frac{63}{392} - \frac{126}{392} - 1 \]
Combine the fractions:
\[ y = \frac{63 - 126}{392} - 1 = \frac{-63}{392} - 1 = \frac{-63}{392} - \frac{392}{392} = \frac{-63 - 392}{392} = \frac{-455}{392} \]
Thus, the vertex of the parabola is at:
\[ \left(-\frac{3}{28}, -\frac{455}{392}\right) \]
So, the ordered pair representing the vertex is:
\[ \left(-\frac{3}{28}, -\frac{455}{392}\right) \]