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 has a horizontal tangent, we first find its derivative and set it equal to zero. 1. The function is \[ y = 6x^2 + 2x - 5. \] 2. Its derivative is \[ y' = \frac{d}{dx}(6x^2) + \frac{d}{dx}(2x) - \frac{d}{dx}(5) = 12x + 2. \] 3. Setting the derivative equal to zero gives \[ 12x + 2 = 0. \] 4. Solving for \( x \): \[ 12x = -2 \quad \Rightarrow \quad x = -\frac{2}{12} = -\frac{1}{6}. \] 5. Substitute \( x = -\frac{1}{6} \) back into the original function to find \( y \): \[ y = 6\left(-\frac{1}{6}\right)^2 + 2\left(-\frac{1}{6}\right) - 5. \] 6. Compute each term: - First term: \[ 6\left(-\frac{1}{6}\right)^2 = 6\left(\frac{1}{36}\right) = \frac{6}{36} = \frac{1}{6}. \] - Second term: \[ 2\left(-\frac{1}{6}\right) = -\frac{2}{6} = -\frac{1}{3}. \] - Third term: \( -5 \). 7. Combine the terms: \[ y = \frac{1}{6} - \frac{1}{3} - 5. \] Write all fractions with a common denominator \(6\): \[ \frac{1}{6} - \frac{2}{6} - \frac{30}{6} = \frac{1 - 2 - 30}{6} = -\frac{31}{6}. \] The point at which there is a horizontal tangent is \[ \left(-\frac{1}{6}, -\frac{31}{6}\right). \]