00:01
Given the horizontal tangents 2 -0 and negative 26, we want to come up with a cubic function, that is a x -cubed plus b -x squared plus c -x plus d, that will give us the two horizontal tangents.
00:16
Now, one way to solve this is through systems of equations, because we have to solve for four different variables at the same time, that is a, b, c, and d.
00:25
In order for systems of equations to work here, though, we're going to need at least four equations that contain a, b, c, and d, as you need at least one equation per variable you want to solve for.
00:38
So one way we can get two of our initial equations is with the horizontal tangents we're given, that is that we know that when we plug in 2, we get out 0, when we plug in negative 2, we get out 6.
00:52
So doing this with our function, if we plug in 2, we will get 8a plus 4b plus 2c plus d and that would be equal to 0.
01:09
And then likewise with negative 2, we will get negative 8a plus 4b minus 2c plus d, and that will be equal to 6.
01:28
So now we have two equations here, but we need at least 4.
01:32
And one of the cool things we can do with these horizontal tangents is one of the defining features of a horizontal tangent.
01:38
Is they happen when the derivative of your function is equal to zero.
01:48
So if we derive our function and set it equal to zero, we can use these same horizontal tangents to come up with two more equations.
01:58
And so deriving our function, using the power rule on all four terms, will give us 3a x squared plus 2b x plus c.
02:11
That is equal to the derivative of our function.
02:14
Function.
02:17
Now if we just take this derived function and plug in 2 and negative 2, that will give us two of our other functions.
02:28
And these will both be equal to 0 because as we said earlier, one of the defining parts of a horizontal tangent is that the derivative of your function should equal 0 at that point.
02:40
Doing this will give us 12 a plus 4b plus c is equal to 0 and doing this with negative 2 give us the same thing but with a negative 4b instead of positive 4b now to solve for these equations what we can notice with the third and the fourth one is that they both have a 12a and they both have a c so if we subtract these two from each other that is to say 12a plus 4b plus c minus 12a minus 4b plus c doing so we'll cancel our a's into 0a we'll make our bs into a positive 4b minus a negative 4b which will be a positive giving us 8b plus c minus c which will give us 0c is equal to 0 and from here if we divide 8 well 0 over 8 is going to give us b is equal to 0 and now that we know that b is equal to 0 we can thin up a lot of our equations because all of them technically no longer have a b term as it's equal to zero.
04:01
If we now see our attention towards the first two equations, we'll see that we have something similar, except with 8a and negative 8a and 2c and negative 2c.
04:13
And that is, they're the exact same constant terms in front of each variable.
04:18
The only difference is that the signs are different...