Question 29

Explanation

Given curve $y = 6\sqrt{x}$ with tangent point $x = 9$

Coordinates of Tangent Point

For $x = 9$, then

The coordinates of the tangent point are $(9, 18)$

Gradient of Tangent Line

Calculate the derivative of the function $f'(x) = m$

Gradient at $x = 9$

Tangent Line Equation

Tangent line equation at point $(9, 18)$ with $m = 1$

Therefore, the equation of the tangent line to the curve is $y = x + 9$