Find the point(s) where a tangent line to the curve \( y=f(x)=3 x^{2} \) has slope \( m_{t a n}=18 \). The point(s) is (are) \( \square \). (Type an ordered pair. Use a comma to separate answers as needed.)

To find the points where the tangent line to the curve \( y = f(x) = 3x^2 \) has a slope of \( m_{tan} = 18 \), we first need to find the derivative of the function, which gives us the slope of the tangent line at any point \( x \). 1. **Find the derivative**: \[ f'(x) = \frac{d}{dx}(3x^2) = 6x \] 2. **Set the derivative equal to the desired slope**: We want to find \( x \) such that: \[ 6x = 18 \] 3. **Solve for \( x \)**: \[ x = \frac{18}{6} = 3 \] 4. **Find the corresponding \( y \) value**: Now we substitute \( x = 3 \) back into the original function to find \( y \): \[ y = f(3) = 3(3^2) = 3 \cdot 9 = 27 \] 5. **Write the point**: The point where the tangent line has a slope of 18 is: \[ (3, 27) \] Thus, the point(s) is (are) \( \boxed{(3, 27)} \).