1. The problem asks to find the gradient of the tangent to the curve $y = x^2 + 3x - 2$ at the point $P(3,16)$. 2. To find the gradient of the tangent to a curve at a point, we use the derivative of the function. The derivative gives the gradient of the tangent line at any point $x$ on the curve. 3. The function is $y = x^2 + 3x - 2$. The derivative, denoted $\frac{dy}{dx}$ or $y'$, is found by differentiating each term: $$\frac{dy}{dx} = 2x + 3$$ 4. To find the gradient at $P(3,16)$, substitute $x=3$ into the derivative: $$\frac{dy}{dx}\bigg|_{x=3} = 2(3) + 3 = 6 + 3 = 9$$ 5. Therefore, the gradient of the tangent to the curve at point $P$ is $9$. This method uses differentiation to find the slope of the tangent line at a specific point on the curve.