00:05
Okay, here's another question that gives us a function, x squared plus bx minus 25, and a maximum value of negative 50.
00:12
It asks us to find a value of b such that the function takes that maximum value.
00:18
In this case, it's actually a minimum value of negative 50.
00:24
Right.
00:24
So recall that the maximum minimum values occur at the x component of the vertex, which is negative b over 2a.
00:32
Right and so a in our case is actually just one so that's going to be for our function negative b over 2a is the same as negative b over 2 right so all we have to do is plug that in to our function right so let's evaluate f of negative b over 2 for x so negative b squared is b squared over 4 right so negative b over 2 squared will be b squared over 4 plus b times uh the that's negative b over 2, which is actually going to be minus b squared over 2, minus 25.
01:09
Now, we also know that the minimum value is negative 50.
01:14
All the ways we do now is solve for b...