Hi, guys.
Welcome to Algebra 1.
Today's lesson focuses on
finding values for elements
in the domain
of quadratic functions.
Your knowledge
of the coordinate plane,
of evaluating expressions
will come in handy
during this lesson.
You ready to get started?
Let's go.
Okay.
To warm you up to
what we're doing here,
I want to throw back a bit
to find y-intercepts
of quadratic functions.
Here we have the function:
f of x equals x squared,
plus x minus 20.
Because of what we know
about quadratic functions,
we know the y-intercept
is typically our constant term
right here,
on the right side.
Here I could say,
"All right, it's negative 20."
What we want to do,
to warm you up to what
we're doing,
is I'll show why it's always
that constant term,
that negative 20, okay?
When we have a y-intercept,
we know that x must be 0
and y is
some numerical value, right?
In order to show
the algebra behind this,
we'll substitute 0 in for x
and see why it is
that we get negative 20.
Okay?
Let's say f of 0--
because we're replacing
that x with 0--
equals 0 squared
plus 0 minus 20, right?
So 0 squared--that's 0--
plus 0 minus 20.
0 plus 0 minus 20
is just negative 20.
We'd say, "Okay,
our y-intercept
is negative 20."
What we did to verify
that our y-intercept
is really going
to be negative 20,
like we thought,
we did that algebra,
and substituted 0 for x.
You can substitute any value
in the domain of a function,
evaluate it at that value,
and see what you get
for the output.
That's what we'll do.
I wanted to warm you up
by showing what I meant
with y-intercepts
in doing the same thing.
Look at this example.
Here you have: f of x equals
x squared plus 5x plus 6,
and you're asked
to find f of 3.
What I'm going to do
is substitute 3
for my x value
and see what I get
for my output.
I'll find "f of 3"--
that's how it's read...

(female narrator)
It's written as f
with a 3 in parenthesis.
...equals 3 squared
plus 5 times 3
plus 6.
Okay?
So, 3 squared--
that's 9--
plus 5 times 3--
that's 15--
plus 6.
9 plus 15--that's 24--
and 24 plus 6 is 30.
I'd say, "Okay,
f of 3 equals 30."
And you'd be all done
with that one.
You'd evaluate
this function at 3
and you saw
its output was 30.
Then you'd read that final
answer as "f of 3 equals 30."
Okay?
Let's try
another one together.
Here's the same function,
f of x equals x squared
plus 5x plus 6,
but now you're asked
to find f of negative 1.
Our process is the same;
it's just our number
that's changed.
Just how we started
the last one.
We're going to substitute
negative 1 for x.
We'll have f of negative 1
equals negative 1 squared,
plus 5 times negative 1,
plus 6.
Now we just want
to evaluate this right side.
So negative 1 squared--
that's positive 1.
5 times negative 1
is negative 5.
And then we have plus 6.
So, 1 minus 5,
that's negative 4,
and negative 4 plus 6
is positive 2.
So we'd say, "Okay,
f of negative 1 is positive 2."
If negative 1's my input
for this function,
then 2 is my output.
That's exactly
what this means.
Good job.
It's time to try
a couple on your own.
Go ahead and press pause.
Take a few minutes and
work through these problems.
To compare your answers
with me, press play.

(female narrator)
Given: f of x equals
x squared minus 3x plus 2.
Letter A:
Find f of negative 2.
Letter B:
Find f of 4.
Let's see what you got.
Let's switch
to our pointer tool
to move these
out of our way.
For f of negative 2,
you should have got 12--
f of negative 2 equals 12.
And for B, f of 4 equals 6.
To see how I got these,
stick with me,
and I'll show you my work.
Let's switch
to my pen here.
For f of negative 2,
what we'll do
is substitute negative 2
for x in our function.
Okay?
So f of negative 2
equals negative 2 squared
minus 3
times negative 2
plus 2.
Right? Okay.
So, negative 2 squared,
that's positive 4.
Negative 3 times negative 2
is positive 6.
And then we have our plus 2.
Then just left to right:
4 plus 6 is 10,
10 plus 2 is 12.
That's how I got
that f of negative 2
was positive 12.
When negative 2 is my input,
12 is my output.
All right?
Now let me show you
how I got B.
Here I was asked
to find f of 4.
So f of 4
equals 4 squared
minus 3 times 4
plus 2.
Okay?
So 4 squared, that's 16.
Negative 3 times 4
is negative 12.
Then we've got our plus 2.
So, 16 minus 12, that's 4,
and then 4 plus 2 is 6.
That's how I got
that f of 4 equals 6.
All right?
When 4 was my input,
6 was my output.
Good job on that.
Let's look at some graphs.
Given the graph
of a quadratic function,
you should be comfortable
with being given an input
and being able
to find the output.
Here you're asked
to find f of negative 1.
I'm going to switch
to my highlighter here.
Okay.
Now, f of negative 1--
so that's an input value.
I need to find negative 1
on my x-axis--
here--on my graph
it's above me at this point.
So I'll trace it up.
There--I land
right there at this point.
I need to find out
the y value right there.
I'm going to trace that
over to the y-axis
to figure out that this
is the point

(negative 1, positive 1).
I know that when negative 1
is my input, 1 is my output.
Okay?
I'd return to my problem
and I'd write that
as "f of negative 1
equals positive 1."
When negative 1's my input,
1 is my output
for this function.
Let's try
another one together,
then I'll let you
chart one on your own.
Now we've got f of 2.
Let me switch
to my highlighter.
I'm going to 2
on my x-axis,
because 2
is an input value here.
At 2,
my parabola is above me,
so I'm going to trace up,
and I land right there.
I'm going to put a point.
Now I'll trace that
over to the y-axis
to figure out
the output.
And it looks like
I am at 1, 2, 3, 4.
When 2 is my input,
4 is my output
because this
is the point (2, 4).
To rewrite it
in function notation,
going back to the form
of our problem,
I'd say that
f of 2 equals 4
because when 2 is my input
for this function,
4 is my output.
Okay? All right?
Time for you to try one.
Press pause,
take a few minutes
and work
through this problem.
You're asked
to find f of 1.
To check your answer
against mine, press play.

(female narrator)
On the graph, the parabola
curves up from
the point of origin.
On one side, the curve
goes to minus 1, plus 1,
and then minus 2, plus 4.
On the other side
of the parabola,
the curve goes to plus 1,
plus 1,
and then plus 2, plus 4.
Find f of 1.
Let's take
a look at how you did.
I'll switch
to my highlighter here.
So f of 1--
1's an input value,
so I'll go
to the x-axis.
At 1, my parabola's above me,
so I'll go up.
I landed right there;
I'm going to put a point.
And I'll go back
to the highlighter tool
and trace it over
to the y-axis
to see where I'm at.
It looks like I'm at 1,
right there.
So when my input value's 1,
my output value is also 1
for this function.
I'd rewrite the answer,
coming back to my problem:
f of 1 equals 1.
When 1 is my input,
1 is also my output
for this function.
All right?
Good job on that!
Great job, guys,
evaluating functions
at values given
from their domain.
I hope your knowledge
of the coordinate plane
and how to
evaluate expressions
really came in handy.
See you soon
for more Algebra 1. Bye!