If you look closely -- and maybe if you’ve been doing a lot of graphing lately --
you’ll start seeing the same symmetrical U-shaped curve all around you.
Stop for a sip at one of those old-fashioned metal water fountains and you’ll be drinking
from the peak of what looks like a tiny parabola
-- a shape first studied in ancient Greek geometry and made by slicing a cone in a particular way
Or take a gander at the path of a jumping fish, grasshopper, kangaroo, or frog -- another parabola.
And the main cable of a suspension bridge is pulled into a parabola by the forces of
gravity and compression.
It’s the same pattern, sometimes bigger or smaller or narrower or fatter.
Not all of them are true parabolas, but an impressive number of the phenomena you notice
will be parabolas,
at least in ideal circumstances.
With so many different manifestations of parabolas out there, it seems like each one could have
a vastly different equation.
But no.
While there’s plenty of unique and complex math underlying the hydraulics of a drinking
fountain or the leap of a kangaroo,
the basic parabolas are all versions of the same thing.
And the same is true for the parabolas you find in college algebra.
They may all look a little different, but each one can be graphed as a variation on a single
quadratic theme.
G’Day, I’m James Tanton, and this is Study Hall Algebra presented by Arizona State University
and Crash Course.
Parabolas were first described by ancient
Greek scholars with words and geometry.
It would take over a thousand more years before algebra was invented and we could describe
them with equations
-- all of which are variations of the simple: y = x^2.
Now no matter what we’re graphing, a sure fire way to get started is to plug some numbers
into the equation as x-values,
and see what matching y-values make the equation true.
Zero squared is zero, so x = 0 and y = 0 make our equation a true sentence.
Then 1 squared is 1, and –1 squared is also 1. 2 and -2 squared are 4. 3 and -3 squared are 9.
We can keep going until we’re sure of the general shape.
Here, we’re pretty sure that if we plug in bigger positive or negative x-values,
we’ll always get bigger y-values.
Here we’re working from an equation, so drawing a line “connecting the dots” is
really just imagining we’ve plugged in all the numbers and plotted all the data.
In this graph, the vertex, or the point where the curve changes direction, is at zero.
It's the low point of the graph, or the minimum.
And that’s it! That’s the graph of the only parabola we’ll ever need.
Well, look at that, we’ve made good time.
Wow, golly gee, it's beautiful outside...
OK, that might be a little hasty.
But jokes aside, the graph of any other quadratic equation is just this plot transformed
-- the basic symmetry always stays the same.
Which is kind of magical! Almost like y = x^2 is a chameleon.
Or maybe there’s a secret mathematical chameleon helping with all the possible transformations.
Well, we can dream.
Before we get started, let’s see what we‘re working with.
Remember a standard quadratic equation looks something like y = ax^2 + bx + c,
where x is our variable and a, b, and c are constants.
So with a = 1 and b and c = 0, y = x^2 is the simplest quadratic relationship out there.
But what happens to the graph if we have an “a” that’s different from 1?
Well, if our graphing chameleon makes “a” equal 1/2, the y-values will only be half
what they were before.
And *snap* the parabola is shallower and wider.
And if she transforms “a” into 2, then the y-values will be twice what they were before
and the graph will look narrower and steeper.
You can think of messing with the “a” coefficient as adjusting the “arms” of the parabola.
Smaller a-values force the arms to lay flatter, while larger a-values bring the arms closer together
All throughout these adjustments, the vertex is still at 0 and stays the lowest point.
Now let’s have her bring back our original y = x^2 graph. What if the “a” term were negative,
like y = –(x^2)? We can probably figure it out before plotting it.
Our y-values are now going to be the opposite of what we had before, so the arms will stretch
down instead of up.
But if we flip our graph, where do we put zero?
Well, negative zero squared is still zero.
So in this graph the vertex is still at zero, but now it’s the high point (the maximum)
rather than the low point (the minimum).
For quadratic equations, the vertex anchors the rest of the curve.
And it’s often the most interesting part of the graph when the math is describing something
in the real world.
It's important to know the minimums and maximums of graphs, which are the lowest and highest
points we've been talking about.
The maximum, for example, can tell us how high the rock launched from our backyard trebuchet
will go.
Now in all of these graphs so far, the vertex has been stuck at the origin where x=0 and y=0.
We’ve only worked with the “a” coefficient so far, and that didn’t shift the vertex
from zero.
So if we’re going for a transformation where the vertex is somewhere else, the “b”
and “c” terms must handle the action here.
There’s actually a way to rewrite quadratic equations to let us understand what the "b"
and "c" terms do to our parabola.
By completing the square like we did when we first explored quadratic equations, we
can give y = ax^2 + bx + c
a shiny new look called vertex form.
The point (h, k) is actually the vertex, and we can read it right off the equation!
If we just have an "a" value, like in y = x^2, y = 2x^2, or y = -(x^2) we can see that h
is 0 and k is 0.
Our vertex for all these parabolas is at the origin, which we've seen so far in all our graphs.
"Could you come up with something a bit more complicated? Oh that’s good."
Now, let’s try converting this trickier equation to vertex form by completing the square.
So x times x is x^2 – that’s straightforward.
We need a –2 and a –2 to add up to –4x, and those multiply to 4.
But our “c” term outside the parentheses is 3, so we’ll need to subtract 1. And we
have y = (x – 2)^2 – 1.
So where is the vertex? h is 2, because we have (x – h)^2,
and the subtraction there is built into the vertex formula – it doesn’t mean that
whatever number h is must be negative.
But because it’s “plus k” in the vertex formula, our k value is negative 1.
(We can think of it as “plus negative one,” which is just –1).
So we got the same answer -- our vertex is still at the point (2, –1).
In fact, by completing the square we see that every quadratic equation really looks like
y = a(x-h)^2 + k in disguise.
They all have the same structure!
And so we’ve confirmed we’ll always get the same symmetrical U-shaped parabola --
it just might’ve been transformed a bit.
In fact, we can put symmetry to work and get a rough sketch of any quadratic equation without
much work at all!
First, let's take the same equation as before.
We can rewrite the equation by pulling out a common factor of x in the first two terms
to get y = x(x-4) + 3.
Now we see that x=0 makes y = 3 and x=4 also makes y=3 --- because both zero and 4 make
the first part 0.
That means there are two points with the same y-value:
(0,3) and (4,3), and we already know the graph is symmetrical.
The line of symmetry must fall halfway between 0 and 4 at x = 2.
So the vertex is somewhere on that line of symmetry and has 2 as an x-value.
But where?
Well, plug in x=2 and we see y = 4 - 8 + 3 = -1. The vertex is at the point (2,-1).
The same place we deduced it was by converting the equation into vertex form!
Now let’s take a look at one of the classic applications of graphing quadratics: projectile motion.
Around the turn of the 17th century, the astronomer and mathematician Galileo
did a lot of work on projectile motion and deduced that all objects fall at the same
rate of acceleration due to gravity,
or about -16 feet per second squared.
This led to the discovery that the heights of objects in motion follow this quadratic formula,
where v is how fast we throw the object up and d is how high we start above the ground.
Now if we’re throwing something, it’s easy to see a parabola as the actual path
of the projectile,
like a ball flying back and forth in an arc.
But most of the time we throw objects at an angle, so their velocity is a mix of vertical
motion -- height changing over time --
and a horizontal motion that spreads those changing heights over time.
And these two directions are a bit harder to model and need more variables,
so a quadratic equation won't be enough.
That’s why in an algebra class things are thrown straight up so we can represent them
with quadratic equations.
The variable on the horizontal-axis for a projectile is time, and the variable on the
vertical-axis is height.
Maybe you want to launch something up that won’t come down – at least, not in the
same form.
Like a firework! It's super important to calculate the maximum height of a firework
because you don’t want it to go off too soon after it’s launched,
and you especially don’t want it to go off when it’s falling back down.
So if you know the launch speed of your firework,
how do you know how long its time-delay fuse should burn before it explodes?
Let’s put on these safety goggles to have a closer look.
Let’s say you and your local fire department have been assigned to set off some celebratory
fireworks from the bridge, over a nearby lake.
To set your fuse properly, you want to find how long it takes the fireworks to reach their
maximum height.
And from Galileo’s work, we know the height of a falling object is h = –16t^2 + v*t + d.
You know the fireworks launch upward at 200 feet per second,
so their initial velocity v is 200.
And the bridge is 75 feet above the water, so the initial height d is 75.
The variables might be different, but this equation is recognizably a quadratic.
So we know its graph is going to be a parabola.
The “a” coefficient here is -16, which means our graph is going to be upside down
and narrower than y = x^2.
And in fact the vertex will be the maximum.
And since “b” and “c” are both nonzero, that vertex will be not be at the origin.
So to sketch a quick graph, we can find two points with the same y-value.
If we write the equation like this, we can notice at time t=0 and t=12.5
the height of the firework above the water is 75ft.
Then since the graph is symmetric,
the fireworks reach their maximum height halfway between those times at t = 6.25 seconds.
Plug that in and our fireworks will launch 700 feet in the air!
We can even learn a bit more from the graph.
If we extend our sketch, the parabola’s arms will cross the horizontal axis when h=0,
or at the roots we discussed in episode 09.
The positive root is somewhere between t= 12 and t=13.
So the graph tells us that your fuse needs to be set so that the fireworks go off
after 6.25 seconds, when they’re 700 feet up, and will be back on Earth just before
13 seconds are up.
Now look at all these quadratics we’ve graphed! A fireworks-worthy accomplishment.
So when setting out on transformations of your own,
remember that all graphs of quadratic equations are essentially graphs of y = x^2
just flattened or widened or shifted around.
And the vertex form of a quadratic equation lets you read of the vertex (h,k) right away.
Remember that h is being subtracted from x, so if you have something like y = (x + 1)^2
+ 3, h must really be negative 1.
But to easily graph a quadratic, use the symmetry of the parabola.
Rewrite it as y = x(ax + b) + c and find the two x values that give the same y-value c.
Halfway between those points is the line of symmetry, and the vertex.
Next time we’ll revisit linear equations and see how graphing can even help us solve
simultaneous equations.
Until then, cheers!
Thanks for watching Study Hall Algebra, which is produced by Arizona State University and
the Crash Course team at Complexly.
If you liked this video and want to keep learning with us here in Study Hall, be sure to subscribe.
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in the links in the description.
See you next time!