To find the vertex of the parabola given by the equation \( y = 14x^2 + 3x - 1 \), we can use the vertex formula for a quadratic function of the form \( y = ax^2 + bx + c \):
The x-coordinate of the vertex is given by the formula: \[ x = -\frac{b}{2a} \]
In your equation:
- \( a = 14 \)
- \( b = 3 \)
Plugging in the values: \[ x = -\frac{3}{2 \cdot 14} = -\frac{3}{28} \]
Now, substitute \( x = -\frac{3}{28} \) back into the equation to find the y-coordinate of the vertex: \[ y = 14\left(-\frac{3}{28}\right)^2 + 3\left(-\frac{3}{28}\right) - 1 \]
Calculating it step by step:
- \( \left(-\frac{3}{28}\right)^2 = \frac{9}{784} \)
- \( 14 \cdot \frac{9}{784} = \frac{126}{784} = \frac{63}{392} \)
- \( 3 \cdot -\frac{3}{28} = -\frac{9}{28} = -\frac{126}{392} \)
- Now, substitute back: \[ y = \frac{63}{392} - \frac{126}{392} - 1 = -\frac{63}{392} - 1 = -\frac{63}{392} - \frac{392}{392} = -\frac{455}{392} \]
The vertex of the parabola is located at the point \((-6, -11)\).
Thus, the correct ordered pair for the vertex is: (−6,−11).
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