A curve has equation $y = x^3 - 48x$ The point A on the curve has x coordinate $-4$ The point B on the curve has x coordinate $-4 + h$ 15 (a) Show that the gradient of the line AB is $h^2 - 12h$. - AQA - A-Level Maths Pure - Question 15 - 2018 - Paper 1

To find the gradient of the line segment AB, we need the coordinates of points A and B.

First, calculate the coordinates of point A:

• At $x = -4$, we find the y-coordinate: $y_A = (-4)^3 - 48(-4) = -64 + 192 = 128$
  Thus, point A is $(-4, 128)$.

Next, calculate the coordinates of point B:

• At $x = -4 + h$, find the y-coordinate: $y_B = (-4 + h)^3 - 48(-4 + h)$ Expanding $(-4 + h)^3$ gives: $(-4+h)^3 = -64 + 48h - 12h^2 + h^3$
  Therefore, $y_B = (-64 + 48h - 12h^2 + h^3) + 192 - 48h$ Simplifying this results in: $y_B = h^3 - 12h + 128$

Now, the gradient ($m$) of line AB can be calculated using the formula: m = rac{y_B - y_A}{x_B - x_A}
Substituting the values: m = rac{(h^3 - 12h + 128) - 128}{(-4 + h) - (-4)} = rac{h^3 - 12h}{h}
This simplifies to: $h^2 - 12$ To find $y_B - y_A = h^3 - 12h$, thus resulting in: $m = h^2 - 12h$.