Find an equation of the tangent to the curve y = (x - 2)⁴ at the point where x = 0 - AQA - A-Level Maths Pure - Question 5 - 2022 - Paper 1

Next, we substitute x = 0 into our derivative to find the slope of the tangent line at that point:

$\frac{dy}{dx}\bigg|_{x=0} = 4(0 - 2)^{3} = 4(-2)^{3} = 4(-8) = -32$

So, the slope of the tangent line at x = 0 is -32.

Now we need to find the y-coordinate when x = 0:

$y = (0 - 2)^{4} = 16$

We have a point (0, 16) on the curve and a slope of -32. Using the point-slope form of the line equation:

$y - y_1 = m(x - x_1)$

Substituting the values, we get:

y - 16 = -32(x - 0)
y = -32x + 16

Thus, the equation of the tangent line is:

y = -32x + 16.