1. The problem is to find the slope of the tangent line to the curve \(y=x^3+4x-7\) at the point where \(x=2\). 2. The slope of the tangent line at any point is given by the derivative of the function evaluated at that point. 3. Differentiate \(y=x^3+4x-7\) with respect to \(x\): $$\frac{dy}{dx} = 3x^2 + 4$$ 4. Substitute \(x=2\) into the derivative to find the slope at that point: $$m = 3(2)^2 + 4 = 3 \times 4 + 4 = 12 + 4 = 16$$ 5. Therefore, the slope of the tangent line to the curve at \(x=2\) is \(16\).