Let's talk about quadratic functions.
One thing we'd like to be able to do is draw the graph of a quadratic function.
Let's look at some features that will help us to draw the graph.
First of all, this is a quadratic function
since we see the x squared
and no other higher powers of x.
We know that the shape of the graph is going to be a parabola,
and we also know the domain is going to be the set of all real numbers.
Perhaps the most important feature for the parabola is its vertex, so let's find that.
We can use this formula: negative b over 2a, f of negative b over 2a.
Of course we need to know what are a and b.
Those are going to come from our quadratic function.
The letter a is the coefficient of x squared,
and b is the coefficient of x.
Let's find negative b over 2a first.
Here, b is negative 2,
so this will be a negative, negative 2
over 2 times a. Here, a is 1.
If we simplify this, this is positive 2 over 2 or 1.
The next thing I need to find then is the y-coordinate of the vertex.
I'm going to do that by using 1, the x value that we just found,
and find f of 1.
I substitute this for the x in the original problem.
1 squared minus 2 times 1 minus 3 equals negative 4.
My vertex is the point (1, -4).
Let's go ahead and put that on our graph.
A second feature we should discuss
is which way the parabola is oriented.
Since the coefficient of x squared is positive,
we know that this parabola should open up.
We can have an idea of the shape of the parabola here,
but we need some more information to make it accurate.
Another important feature of any graph would be intercepts.
Let's see if we have any.
We'll start with the y-intercept.
The y-intercept is the y-coordinate of any point
where the graph touches or crosses the y-axis.
To find it, we let x equal 0.
We need to find here f of 0, which is negative 3.
There's my y-intercept at (0, -3).
Let's plot the point. Let's look for x-intercepts.
The x-intercept is the x-coordinate of any point
where the graph crosses the x-axis.
To find them, let y equal 0.
Solve for x. Substitute 0 for y or f of x.
This one we can factor.
We want x minus 3 times x plus 1.
This gives me two x-intercepts: 3 and negative 1.
I can plot these points on my graph.
Now remember the graph is the shape of a parabola,
so now I'm going to connect these
with a smooth, parabola-shaped curve.
This is my function f. Now that we have the graph,
there are many other features that we can discover
about this function.
Now that we've graphed this quadratic function,
let's look at other features.
First of all, we notice that our graph has some symmetry.
We can draw in the line of symmetry for our graph.
It is a vertical line passing through the vertex.
The equation of a vertical line is always
x equals the x-intercept. This is x equals 1.
We can also find the range.
The range is values that y is allowed to have.
If we look at this, we see that the lowest y value occurs at the vertex,
and the y value there is negative 4.
Anything larger than negative 4 is also allowed.
There is one final question we want to answer.
On what intervals is the function increasing or decreasing?
We want to look from left to right
as we answer this question.
Starting at the vertex, as the x values get larger,
notice that the function values are rising.
They're increasing. f is increasing starting at the vertex,
which has an x-coordinate of 1,
and as x gets larger without bound.
On this portion of the graph,
we see that the y values are getting smaller
as x gets closer to the vertex.
The x-coordinate of the vertex is 1,
so f is decreasing on negative infinity to 1.
Let's look at another example.
Here's another quadratic function. Let's look at its features.
The first thing I notice here
is that the coefficient of x squared is a negative number.
That tells me this parabola is going to open down.
Another thing I should write down right away
is the domain of this quadratic function.
The domain is all real numbers. Let's find the vertex.
We're going to proceed by finding the x-coordinate of the vertex:
negative b over 2a.
Remember that b is the coefficient of x,
and a is the coefficient of the x squared term.
If we simplify this, 6 over negative 6 is negative 1.
To find the y-coordinate of the vertex,
I need to find f of negative 1.
If we work this out, this is negative 3 plus 6 minus 2, which is 1.
Now I can go back and write the coordinates of the vertex,
which are (-1, 1).
Let's plot that on our graph.
Now we've already said that this parabola will open down,
so let's picture the parabola on our graph.
It opens down, so I can see it's going to eventually cross the y-axis,
and it's going to cross the x-axis twice.
I know I need to find the intercepts.
To find the y-intercept, we let x equal 0.
We want to find f of 0. That's going to be 0 minus 0 minus 2.
That's negative 2.
Let's put our y-intercept on our graph: (0, -2).
Now let's look for our x-intercepts.
To find an x-intercept, you let y equal 0.
We're going to substitute 0 for y or f of x,
and we're going to try to solve this equation.
Now I'd like to factor this, but it doesn't factor easily.
Let's use the quadratic formula:
x must be equal to negative b, which will be a negative, negative 6,
plus or minus the square root of
b squared, which is negative 6 squared, minus 4
times a, which is negative 3, times c, which is negative 2,
and all of this is over 2 times a,
which we said is negative 3. Okay, now simplify this.
This is positive 6 plus or minus the square root of
negative 6 squared, which is 36,
and here I have negative 4 times negative 3 times negative 2.
That's three negative numbers, so the result is negative.
4 times 3 is 12 times 2 is 24.
2 times negative 3 in the denominator is negative 6.
This is 6 plus or minus the square root of 12 over negative 6.
We want the x-intercepts in order to draw the graph.
We need to know approximately where to put them on the graph,
so let's use our calculator and approximate the x-intercepts.
The first one is open parentheses 6 plus
the square root of 12.
Close your parentheses, and then divide that by negative 6.
This is approximately negative 1.58. Now let's find the other one.
Again, open parentheses 6 minus the square root of 12.
Close your parentheses, and then divide this by negative 6.
This one is approximately negative 0.42.
We're not going to be able to locate them precisely on our graph,
but we can get them in the right area. Negative 1.58 is between negative 1 and negative 2,
and negative 0.42 would be approximately here.
Remember that the shape is a parabola that opens downward.
You want to make sure this is a nice smooth curve.
This is the graph of f. Now let's look at some other features of this graph.
The axis of symmetry is a vertical line passing through the vertex of the parabola.
Let's sketch that on our graph first.
Remember the equation of this vertical line is x equals the x-value of the vertex.
The axis of symmetry is the vertical line x equals negative 1.
Let's find the range. The range is the set of values y is allowed to have.
We see on this function that it has y-values all the way up to and including positive 1.
We'll write the interval from negative infinity to positive 1
and close it with a bracket. The last question we like to answer
is the intervals on which the function f is increasing
and those on which it's decreasing.
Starting on the lefthand side,
notice that the function values are rising
until you get to the vertex.
We say that f is increasing on the interval
from negative infinity to the x-coordinate of the vertex.
That would be negative 1.
At the vertex, the graph turns,
and the function values begin to decrease.
We say that f is decreasing from negative 1 to positive infinity.
Now let's look at some other examples of quadratic functions.
This next example gives us the graph of a quadratic function
and asks us to write the function algebraically.
It's worth noting here that, up until this point,
we've always looked at quadratic functions
in the form f of x equals a times x squared plus b times x plus c where a is not 0.
There is another form of the quadratic function,
and that's called the standard form.
In the standard form, the nice thing is that we can see the vertex
of the quadratic function: h and k.
Then we have this one other value a that we're going to have to find.
In the given graph, we can pick out the vertex.
The vertex has coordinates (2, 1).
We can immediately substitute those values for h and k.
The letter h is 2, and k is 1.
As you can see, we're almost there.
We just have that one missing value, a, that we need to find.
In order to find it, we're going to use the one other given point on our graph.
The function passes through the point (1, 3).
This is an x and a y value that correspond to one another
and make this a true statement.
Remember that f of x is just another name for y,
so we're going to replace f of x with 3.
Then we want to replace that x with 1,
the coordinate of that given point. Now let's solve for a.
We need to subtract 1 from both sides
and get 3 minus 1, which is 2.
Negative 1 squared is 1, so we find out that a is 2.
We can make this substitution here for a,
and we find that our quadratic function is 2
times x minus 2 squared plus 1.
This is in standard form, and, sometimes, that's enough.
Often, you're going to be asked to go ahead and go a little farther
and write this in the other form:
f of x equals a times x squared plus b times x plus c where a is not 0.
In order to do that, we just need to multiply this out
and get rid of the parentheses. Remember the order of operations.
You have to do the squaring before you can distribute the 2.
This is x minus 2 times itself.
That's x squared minus 4x plus 4 plus 1.
Now we can distribute.
The last step is to combine those last two numbers.
Here we have two versions of the equation for this quadratic function:
the standard form and the a times x squared plus b times x plus c form.
This example says, without graphing,
to determine whether the quadratic function has a maximum or minimum value
and then to tell what that value is.
We don't want to draw the graph,
but we probably should think about it a little bit.
This quadratic function has a positive coefficient of x squared.
That means that the graph is a parabola that opens up.
I'm going to think about that parabola.
The vertex is the lowest point on the graph,
so I know this one must have a minimum value.
I also know that the minimum value is
the y-coordinate of the vertex, and that's what I need to find. Okay.
We know how to find the coordinates of the vertex.
Let's do it. To find the x-coordinate, we want to find
negative b over 2a. Here, b is 12,
so we get negative 12 over 2 times 2. Negative 12 over 4 is negative 3.
The next step is to take this and find f of negative 3.
2 times negative 3 squared plus 12 times negative 3 minus 1.
2 times 9 is 18, minus 36, minus 1.
This works out to be negative 19.
That's the y-coordinate of the vertex,
and for this quadratic function it is the minimum value.
Now we have one more application.
The price p and quantity x sold
of a certain product obey the demand equation:
p equals negative 1/6 x plus 80 for x between 0 and 480.
The first thing we want to do is express the revenue as a function of x.
We need to remember that revenue is price times quantity.
If we're writing revenue as a function of x,
price times quantity becomes p times x.
Price is negative 1/6 x plus 80.
Multiply times the quantity x. We're going to distribute the x.
The next thing we want to do is find the quantity x that maximizes our revenue,
what that maximum revenue is, and the price that should be charged
to get the maximum revenue.
We look at our revenue function. This is a quadratic function.
The coefficient of x squared is negative,
so we know the graph is going to be a parabola
that opens down. Let's think about that shape.
This function does in fact have a maximum value.
It's going to occur at the vertex.
We need to find the x-coordinate of the vertex
to answer the first part.
The maximum revenue is the y-coordinate of the vertex.
Then we'll worry about finding the price.
To find the x-coordinate of the vertex,
we want to do negative b over 2a.
That's going to be negative 80
over 2 times negative 1/6.
This is going to end up being a positive number,
and it's 80 divided by 1/3. That's actually 240.
The next thing: Find the revenue when x is 240.
R of 240 is negative 1/6 times 240 squared plus 80
times 240. We need to work this out on our calculator.
Now with our calculator, let's work this out.
For negative 1/6, I'm going to use the fraction key on the calculator.
Times 240 squared plus 80 times 240.
This tells me the maximum revenue is 9,600 dollars.
There's one more part to answer.
We want to know the price that should be charged
to maximize the revenue.
We're going to go back to our price demand equation.
We know that when x is 240, we get the maximum revenue,
so we're going to find the price when x is 240.
Again, we'll use our calculator,
or maybe we don't have to. 1/6 of 240 is 40, so we have negative 40 plus 80
is 40. The company should charge 40 dollars for each product.
This is it for quadratic functions.
Now you need to review these problems
and continue practicing to gain skill and mastery.