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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.