Question: Consider the function $f(x) = x^3$. By first finding the gradient of the secant through $P(2,8)$ and $Q(2+h,(2+h)^3)$, determine the gradient of the tangent to the curve at the point $(2,8)$. $$m = \frac{(2+h)^3 - 8}{(2+h) - 2}$$
1. **State the problem:** We want to find the gradient of the tangent to the curve $y = x^3$ at the point $(2,8)$ by first calculating the gradient of the secant line through points $P(2,8)$ and $Q(2+h,(2+h)^3)$.
2. **Formula for gradient of secant:** The gradient $m$ of the secant line between points $P(x_1,y_1)$ and $Q(x_2,y_2)$ is given by
$$m = \frac{y_2 - y_1}{x_2 - x_1}$$
Here, $x_1 = 2$, $y_1 = 8$, $x_2 = 2+h$, and $y_2 = (2+h)^3$.
3. **Substitute values:**
$$m = \frac{(2+h)^3 - 8}{(2+h) - 2} = \frac{(2+h)^3 - 8}{h}$$
4. **Expand the numerator:**
$$(2+h)^3 = (2)^3 + 3(2)^2h + 3(2)h^2 + h^3 = 8 + 12h + 6h^2 + h^3$$
So,
$$m = \frac{8 + 12h + 6h^2 + h^3 - 8}{h} = \frac{12h + 6h^2 + h^3}{h}$$
5. **Simplify by canceling $h$:**
$$m = \frac{\cancel{h}(12 + 6h + h^2)}{\cancel{h}} = 12 + 6h + h^2$$
6. **Find the gradient of the tangent:** The gradient of the tangent at $x=2$ is the limit of the secant gradient as $h \to 0$:
$$m_{tangent} = \lim_{h \to 0} (12 + 6h + h^2) = 12 + 6(0) + 0^2 = 12$$
**Final answer:** The gradient of the tangent to the curve $y = x^3$ at the point $(2,8)$ is $\boxed{12}$.