00:01
We want to find equation of the tangent, this line, through this point b.
00:10
So i let b be this point.
00:13
I let fx be the function.
00:16
And i let a be the point on this function such that this tangent occurs.
00:25
And i let this point be a.
00:26
So the y value will be fa.
00:30
Let's find our f prime first.
00:34
So f prime will be 3x square minus 9.
00:39
Now we know that at a, where x equals to a, the f prime a will be the same as gradient of ab.
00:55
So f prime a will be 3a square minus 9.
00:59
Gradient of ab will be fa minus 9 over a minus 1.
01:09
So we have 3a square minus 9, a minus 1.
01:16
Fa is a cube minus 9a plus 9.
01:21
So this works out to be as you expand and you work it out, you will get this 2a cube minus 3 a square equals to 0.
01:31
So i'm going to take factorize out the a square.
01:36
I'll get 2a minus 3 equals 0.
01:39
So i actually have two points...