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 22 - 2013 - Paper 1

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

rac{dy}{dx} = 5(2x - 3)^4 imes 2 = 10(2x - 3)^4

Next, we evaluate the derivative at x = w = 0.5:

rac{dy}{dx} igg|_{x=0.5} = 10(2(0.5) - 3)^4 = 10(-2)^4 = 10(16) = 160

Now that we have the gradient (m), we can write the equation of the tangent line in point-slope form:

Using the point (0.5, -32), the equation is given by:

y - (-32) = 160(x - 0.5) \

This simplifies to:

y + 32 = 160x - 80 \

Rearranging gives the final tangent line equation:

$$y = 160x - 112$$