Differentiation and integration » Tangent line with a given gradient  

Below you can see an example in which is shown how you can calculate points on a graph of which the gradient is given.

Given is function f (x) = 13x³ – x² – 2x + 8.
In points A and B on the graph of f is the gradient of the tangent line equal to 1.

Calculate the coordinates of A and B algebraically.

First calculate the derivative.
f ' (x) = x² – 2x – 2

Make the equation with which you can calculate the x-coordinates of A and B and solve it. In other words, you have to solve f ' (x) = 1.
x² – 2x – 2 = 1
x² – 2x – 3 = 0
(x + 1)(x – 3) = 0
x = –1 or x = 3

Now calculate the y-coordinates of points A and B.
Take the smallest of the x-values you just calculated for point A.
f (–1) = 13 · (–1)³ – (–1)² – 2 · (–1) + 8 = 823
f (3) = 13 · 3³ – 3² – 2 · 3 + 8 = 2

So the coordinates are A(–1, 823) and B(3, 2).