Hi, in this video I'm going to
take you through a fairly
standard type of word
problem that
involves quadratic functions.
Though I shouldn't really call
it a word problem because
you're not going to see a
whole lot of words here.
It's more an application, sort
of a real-world use of
parabolas and quadratic
functions.
So what I have here
is a building in a
lovely green field.
And let's say, for the sake of
argument, that you're in this
building and apparently a
little bored because you
decide to entertain yourself
by leaning out a window and
throwing a ball as high
up into the air as
you possibly can.
So let's say you're here
in this window.
You go ahead, you take your
ball, you throw it up in the
air as high as you can.
It's going to reach a maximum
point, turn around, and start
falling down back to Earth,
coming down like this.
So if you look at the shape that
I've just drawn, let's
say your ball is right around
here right now in the process.
You notice I have something
that looks
a lot like a parabola.
And parabolas, the graph
for a parabola goes
with a quadratic function.
Any kind of problem that
involves throwing something,
dropping something, kicking
something, anything like that,
is going to have a shape
that's like a parabola.
So these problems are fairly
common, as I said, word
problems for this
kind of topic.
So let's go ahead and sort of
draw a coordinate axes, an x-
and y-axis in here, to sort of
see how this problem sort of
takes shape.
So first off, our y-axis.
The y-axis is going to
represent where we
started this ball.
So it's going to be
approximately here.
That's my y.
The x is going to represent
the ground.
So here, we're talking
right around here.
This is my x.
And the parabola is the
graph that we have on
this xy axis or axes.
Each parabola has sort of
defining key characteristics
that create or form
this shape.
First off, right here,
the y-intercept.
That's the initial value, the
starting point, the height of
the ball when you threw it.
Next up, we have that maximum
point, where it gets to up
here at the top.
We call that the vertex.
So let me put that in.
Next up, here where it hit
the ground, which is our
x-intercept.
It's a little hard to see
it with the grass here.
Put my line in.
This is our x-intercept.
And for these problems, usually
the concept that
matches the x-intercept
is the idea of,
again, hitting the ground.
Now, for this problem I'm not
going to ask where we started,
when it finished, where it hit
the ground, or what the
maximum is, none of
those things.
What I want is a specific
height.
I want to know when this
ball reached 200 feet.
So first off, we need the
equation, the model
that goes with this.
As I said, it's going to be
a quadratic function.
What we're going to have is h
of t, the height of the ball
at time t, be equal to negative
16 t squared plus
112t plus 40.
And as I said before, what I
want to know is, when does
this ball get to a
height of 200?
Those are both zeroes,
by the way.
So 200 feet.
Since I gave you the height of
the ball, not the time, we're
going to actually use that to
represent or to replace the h.
So instead of just plugging in
200, I'm going to set this
entire equation equal to 200.
That's going to give me 200
equals negative 16t squared
plus 112t plus 40.
And that's the equation
I need to solve.
When I find t, I find my time.
This is a quadratic equation.
You notice we have a t squared
here, and that's the highest
degree term.
For quadratic equations, , we
have a number of different
ways to solve it.
But most of them
involve getting
everything to one side.
Meaning all the terms
together.
So what I'm going to do to help
with that is I'm going to
go ahead and subtract
200 from both sides.
That way, I get this
equal to 0.
That step is critical for most
quadratic equations.
Now, I need to go ahead and pan
up here, because I have
run out of room.
So our drawing is going to go
away for a little bit, but
we'll bring it back later.
The equation that we have now
after we've done this is 0,
which is what we wanted, equal
to negative 16t squared plus
112t minus 160.
Now we need to solve this.
And there are a lot
of different
techniques we can use.
But generally speaking, the
easiest technique, if you can
do it, is to factor.
Now, I don't know about you,
but I'm not really keen on
trying to factor this the
way it is right now.
These numbers are pretty big and
it's going to be difficult
to try to do it.
So the first thing I want to
do is take out a greatest
common factor.
If I can factor out some of
the numbers to make them
smaller, it's going to be a
lot easier to deal with.
And in this example, the
greatest common factor is
actually negative 16.
So if I factor that out, this
quadratic becomes a lot nicer
to deal with.
Doing that, I'm going to get
t squared minus 7t plus 10.
Again, a much nicer kind of
thing to actually factor.
Looking at this, we want two
numbers that multiply to
positive 10 but add
up to negative 7.
The two that are going to work
negative 2 and negative 5.
So I get 0 equals negative 16.
When I factor that out,
it doesn't go away.
We need to include it there.
Times t minus 2 times
t minus 5.
Using the zero product rule,
what we can say is each of
these potentially
could equal 0.
That's the way that we
would get the whole
thing equal to 0.
Well, there's no point setting
negative 16 to 0, it just
doesn't work.
But our next one,
t minus 2, that
potentially could equal 0.
So we'll do that.
We have t minus 2 equals 0.
And our next one, t minus 5,
that could also equals 0.
If either of those are
the case, the whole
thing equals 0.
So this first one, t minus 2
equals 0 when t equals 2.
And here, this one equals
0 when t equals 5.
So after solving our equation,
what we end up with are two
different answers, t equals
2 and t equals 5.
Let's go back to our drawing for
a second and look at what
this means in terms of the
context of the problem.
So back up here.
We get the following answers.
Again, we had t equals
2, t equals 5.
And for many word problems,
sometimes only one answer out
of the two is valid.
But here, we actually have
both answers being valid.
The idea being that we threw
this ball up into the air and
wanted to see when it got
to a height of 200 feet.
Well, if I draw that height and
say this was my 200 feet
right there, you notice the ball
went up in the air until
it got to 200 feet.
Then it kept going all the way
up to its vertex, came back
down, dropped, got to 200 feet
again, and then continued
dropping all the way
down to the ground.
So our final answer for this
problem is that the ball
reaches a height of 200 feet
at two different times.
Let me switch my
pen color here.
Again, it reaches this height
at two different times.
At 2 seconds, meaning
2 seconds after
the ball was thrown.
And again, at 5 seconds,
5 seconds after
the ball was thrown.
Those two times are
our answer.
So this drawing was
not very to scale.
Again, it was just an
illustration to help you see
what was going on.
But let me show you a more exact
sort of mathematical
graph of this exact
same equation.
In this next slide right here,
we have the actual nice formal
graph of that same function
we were looking at
just a second ago.
This is the graph of h of t
equals negative 16t squared
plus 112t plus 40.
So you'll notice for the parts
of this graph, we're going to
start right here at
a height of 40.
The y-intercept is our
initial value.
The ball goes up in the air
until it gets to this height
we're looking for.
The red line here represents
200 feet.
You'll notice the 200 on
the y-axis right there.
The ball then continues up in
the air a little bit higher,
until it gets to its maximum,
its vertex.
Then it comes back down here,
again, to 200 feet.
And if you look closely, this
point right here that we were
looking at, the y is 200.
But down here, the x
is at 2 seconds.
So when x is 2, or in this
instance t is 2,
the height is 200.
Ball goes up to its
height, comes back
down, gets to 200 again.
And if we go ahead and look at
this down on our x-axis, you
notice that that's also at x
equals 5, or here t equals 5.
So those are the two times
that we get that value.
So that's how this works, how
these parabolas work, and how
you can solve applications
with them.
