00:01
So we have an equation that represents the outdoor temperature, and it can be represented this way.
00:08
Y equals f of t, where t is time in elapsed time in hours, and that's going to be equal to negative 0 .3t squared plus 8 .7t minus 2.
00:32
And this is valid for t between 4 and 22 hours elapsed time.
00:44
Again, t is in hours, and y is degrees, temperature, that is.
00:58
The first question is to find the maximum temperature predicted by the quadratic model and the time at which it occurs.
01:06
Well, since this is a quadratic, we note that the line of symmetry for any quadratic, and we do see that this right here is an upside down quadratic, meaning it opens down.
01:20
The vertex for that is going to be at the location minus b over 2a.
01:25
If we think about the standard form of a quadratic, it's y equals a.
01:31
We could do it in x or t is the independent variable, a x squared plus bx plus c, and the line of symmetry is at x.
01:44
Equal minus b over 2a.
01:48
We can use that to find the line in symmetry and that line of symmetry will give us the vertex, that x location right there.
02:02
So in this case, our independent variable is t.
02:05
So this line of symmetry and the vertex is going to occur at t equal to minus b over 2a, which is minus 8 .7 over two times negative 0 .3.
02:23
If you punch it into the calculator, that's going to come out to 14 .5 hours.
02:30
That implies 14 .5 hours t is going to be 30 p .m.
02:42
Because it's 14 and a half hours after midnight...