Find the point \( (x, y) \) at which the graph of \( y=6 x^{2}+2 x-5 \) The function \( y=6 x^{2}+2 x-5 \) has a horizontal tangent lind (Type an ordered pair. Simplify your answer.)

To find the point where the function \( y = 6x^2 + 2x - 5 \) has a horizontal tangent, we first need to compute the derivative of the function. The derivative \( y' \) gives us the slope of the tangent line. Calculating the derivative: \[ y' = \frac{d}{dx}(6x^2 + 2x - 5) = 12x + 2 \] Next, we set the derivative equal to zero to find the critical points: \[ 12x + 2 = 0 \] Solving for \( x \): \[ 12x = -2 \\ x = -\frac{1}{6} \] Now, we substitute \( x = -\frac{1}{6} \) back into the original equation to find the corresponding \( y \): \[ y = 6\left(-\frac{1}{6}\right)^2 + 2\left(-\frac{1}{6}\right) - 5 \\ y = 6\left(\frac{1}{36}\right) - \frac{2}{6} - 5 \\ y = \frac{1}{6} - \frac{1}{3} - 5 \\ y = \frac{1}{6} - \frac{2}{6} - \frac{30}{6} \\ y = \frac{1 - 2 - 30}{6} = \frac{-31}{6} \] Thus, the point at which the graph has a horizontal tangent line is: \[ \left(-\frac{1}{6}, -\frac{31}{6}\right) \]