The curve C has equation y = (2x - 3)^{5} The point P lies on C and has coordinates (w, -32) - Edexcel - A-Level Maths Pure - Question 23 - 2013 - Paper 1

To find the equation of the tangent at point P, we first need to differentiate the curve:

$y = (2x - 3)^{5}$

Using the chain rule:

$\frac{dy}{dx} = 5(2x - 3)^{4} \cdot 2 = 10(2x - 3)^{4}$

Now we evaluate the derivative at x = \frac{1}{2}:

$\frac{dy}{dx} \Big|_{x=\frac{1}{2}} = 10(2(\frac{1}{2}) - 3)^{4} = 10(-2)^{4} = 10 \cdot 16 = 160$

Thus, the gradient of the tangent at P is 160. Now, we can write the tangent equation using point-slope form:

$y - (-32) = 160(x - \frac{1}{2})$

Simplifying this gives:

$y + 32 = 160x - 80$ $y = 160x - 112$

Therefore, the equation of the tangent is:

$y = 160x - 112$