Finding the Minimum or Maximum of Quadratic Functions

Uploaded by MATHRoberg on 14.11.2010

Hi, I'm Kendall Roberg, and today we are going to talk about minimums and maximums.
Let's start by looking at this list of red numbers.
If we were going to try to find the minimum and the maximum number in this list, we would first look through and try to find the smallest number.
That will tell us what the minimum is.
So looking through this list, it looks like -7 is the smallest number.
So we call that the minimum.
Now to find the maximum number, we're going to look through the list again and try to find the greatest number, which looks like 21 is the greatest number. 0:00:40.000,0:00.44.000 So we call that the maximum.
Now we can also find the minimum and maximum of graphs of functions.
Here, we have two different parabolas graphed.
and by looking at this first one, I see that the minimum value, or the lowest value on the graph, is right here.
so let me write "minimum."
Now to find the maximum on this first graph, it's actually kind of difficlut.
I mean, I see it's really tall right here...
but I know those arrows indicate that it keeps going up.
So actually, this has no maximum, it just keeps going up and up and up.
It's difficult for us to find the greatest number, because it keeps going.
But, I am sure that this is the minimum.
Now, if you remember some stuff about parabolas, this is also called the vertex.
So, on this parabola, the vertex is where the minimum value of the function is.
Now, let's look at this one.
On this one, we have a different situation.
I can see that this, right up here, is the maximum value of the function.
So I'll write "maximum."
And again, the maximum is located at the vertex.
We can't identify a minimum value because those arrows indicate that the function keeps going down, both here and here.
What's interesting is that the minimum value was at the vertex on this parabola, and over here, the maximum value was on the vertex here.
Now, I say minimum and maximum, and I'm saying value too.
Well, if we were going to try to find out what that actual value was instead of just pointing to it on the graph, we would actually go to the vertex and draw a line over to the y-axis.
Whatever value this is on the y-axis would actually be the minimum value.
Over here, what ever value this lined up with on the y-axis would atually be the maximum value of the function.
So let's try doing this with a few examples.
Our instructions are to find the minimum or maximum value of two different parabolas.
This first parabola is in standard form.
That is, we have a function in ax^2 + bx + C form.
And if you remember standard form, we can find the vertex of the function by first finding what -b/2a is.
That would give us the x-coordinate of the vertex.
So, b, or the coefficient of the x term, is -4.
So negative -4 is actually just 4.
And 2a...well our a is -1 so 2 times -1 is -2.
So it looks like we have when we reduce...the x-coordinate of the vertex is at -2.
So let's see..1, 2.
The vertex is going to be somewhere at the x = -2 line.
Let's find out exactly where.
To find that out, we're going to plug our -2 into the function and we get...well f(-2) is...let's plug in -2 for the x^2 there.
I think we're going to kind of run out of room here...
Let's erase a little bit...
f(-2) is...well (-2)^2 is 4, times negative is -4, so we get -4 + 8 -5.
Well, -4 + 8 is 4, minus 5 brings us to -1.
So that's our y-coordinate of the vertex.
So it looks like our vertex is going to be right there.
Also, by looking at the coefficient of the x^2 term, we see that this parabola points down.
So here's a rough graph of what this function looks like.
Now, my goal was to find the minimum or maximum value of the function.
I see that because the parabola is pointing down, the maximum value is located at the vertex, right there.
That's the maximum.
The function never has a greater value than it does right at the vertex.
And the vertex was located at (-2,-1), so the maximum value of the function is -1, which is the y-coordinate corresponding to the vertex.
Let's do one more example.
Over here, we have a function that's in intercept form.
So actually, this (x-4) and this (x+2) tells us the x-intercepts are at...we can set each of them equal to 0...and we see that the x-intercepts are equal to 4 and -2.
And we also know that if we were to foil this out, we would have a positive coefficient for the x^2 term.
So I know the parabola points up.
Now, looking at where the intercepts are and knowing it points up, I know this must have a shape like this.
So at this point I can tell this is going to have a minimum value.
Now how do we find that minimum value?
I'm going to leave that up to you. Good luck!