### Video Transcript

In this video, weβre gonna look at
some graphs of quadratic functions and find links between what they look like, and
the various coefficients in their equations. Weβll explore why those links
exist, by considering the quadratic formula.

Remember, the quadratic equation
has got an π₯ squared term, an π₯-term, and a constant term. So thatβs some number times π₯
squared plus or minus some number times π₯ plus or minus some constant number on the
end.

So for example π¦ equals three π₯
squared plus two π₯ minus five. In that case, the π-value is
positive, itβs positive three. The π-value is positive, itβs
positive two. But the π-value was negative, it
was negative five.

Another example is π¦ equals
negative two π₯ squared. So in this case, the π-value, the
coefficient of π₯ squared, is negative two. But the coefficients of the π₯-term
and the constant are just series of π equals zero and π equals zero.

And one more example, π¦ equals a
quarter π₯ squared plus two-fifths. In that case, the π-value will be
a quarter, the π-value will be zero, and the π-value will be two-fifths.

So π or π could be equal to zero,
but to be a quadratic equation, the value of π could never be zero.

If we plot any quadratic function
on a graph, weβll always get a symmetrical parabola like one of these two, either a
U-shape or an upside-downed U-shape. Now you probably remember that the
value of π, the coefficient of π₯ squared that is, tells you about how wide or thin
the curve is, and which way up it is, remember, positive happy smiley curves or
negative sad down-faced curves. And changing the π₯-coefficient, so
the value of π moves the curve left or right on the graph. And the π-value, the constant
term, tells you the value of the function when the π₯-input is zero. In other words, where it cuts the
π¦-axis or the π¦-intercept.

So looking at the coefficients and
the constant term can tell us a lot about what the graph of the function would look
like. And the other way around, if we see
the graph, we can tell what the coe- some of the coefficients are gonna be. But thereβs also another aspect we
need to know about. Where does the curve cut the
π₯-axis? In other words, which π₯-inputs
generate π¦-outputs of zero?

Now you probably spent quite a long
time working out such things, maybe by reading values from graphs, so using
systematic trial and improvement. Perhaps factoring, or using the
quadratic formula, or even by completing the square. But you may have noticed that
sometimes you get two answers, sometimes you get one, and sometimes you donβt get
any. Maybe the quadratic expression
canβt be factored, or maybe the formula just goes wrong and says math error on your
calculator, when you type it in.

Now some quadratics have two roots
and thatβs because they cut the π₯-axis in two places. So there are two π₯-values that
generate π¦-coordinates of zero. Some have one root. For what we call repeated roots,
thatβs two roots, but they just happened to be in the same place. And others have no roots, well no
real roots. There is a way of using things
called imaginary or complex numbers to generate nonreal roots. But weβre not going to worry about
that yet. So if the curve turns around and
heads back up, or maybe turns around and heads back down depending on whether
theyβre coming from above or below, without crossing the π₯-axis anywhere, then we
say there are no roots. And thatβs because there are no
points on that curve that have a π¦-coordinate of zero. None of the π₯-inputs generate a
π¦-coordinate of zero.

Letβs have a good look at that
quadratic formula then. The solutions of ππ₯ squared plus
ππ₯ plus π equal zero, where π is not equal to zero, are given by π₯ is equal to
negative π plus or minus the square root of π squared minus four ππ all over two
π.

So the idea is you take your
quadratic equation and you just plug in the π, π and π values. And thatβll tell you the
π₯-coordinates for which the π¦-coordinate is zero. But whether there are two
solutions, one solution, or no solutions, all hinges on this little bit here; and
itβs called the discriminant. So when using the quadratic
formula, we need to find the square root of the discriminant in order to help us
find the π₯-coordinates of the points where the curve cuts the π₯-axis. Now thatβs the issue. So if the discriminant is positive,
weβre gonna be finding the square root of a positive number. And that will give us two different
values, a positive version and a negative version, so will generate two values of
π₯. If the discriminant is equal to
zero, then weβre gonna be taking the square root of zero, which is zero. So thatβs only one value, so weβre
gonna find just one value. And if the discriminant is
negative, weβre gonna be trying to find the square root of a negative number, which
you know thatβs gonna be very difficult indeed, if not impossible; unless you invent
an entire new system of numbers called imaginary numbers.

Letβs look at a couple of examples
then. So if π¦ equals π₯ squared plus
three π₯ plus two, that means π is one, π is three and π is two. So if we put the π¦-coordinate
equal to zero so that we can find out where it cuts the π₯-axis, if we plug all
those numbers into the quadratic formula, then the discriminant here is three
squared. Thatβs nine minus four times one
times two; thatβs eight. So nine minus eight is positive
one. And the square root of one could be
positive one, could be negative one. Positive one times positive one is
one, negative one times negative one is also one. So that generates two possible
solutions. So for this particular quadratic,
an π₯-coordinate of negative one generates a π¦-coordinate of zero, or an
π₯-coordinate of negative two generates a π¦-coordinate of zero. In other words, it cuts the π₯-axis
in two places when π₯ is equal to negative one and when π₯ is equal to negative
two. So when the discriminant π squared
minus four ππ was bigger than zero, we had two real roots to our equations; two
π₯-values that generate a π¦-coordinate of zero.

Okay. Letβs look at another example
then. π¦ equals π₯ squared plus two π₯
plus one. So in this quadratic π is equal to
one, π is equal to two, and π is equal to one. So if we put π¦ equal to zero to
try to find the π₯-coordinates where it cuts the π₯-axis, and when we plug those
numbers into our quadratic equation, this bit inside the square root here, the
discriminant, turns out to be four minus four, which is zero. So when we go on to solve that
equation, the values of π₯ are gonna be negative two plus or minus the square root
of zero. So obviously the square root of
zero is zero. So weβre adding zero to negative
two and weβre subtracting zero from negative two. So clearly, our two solutions are
gonna be exactly the same, in this case, negative one. So basically, an π₯-coordinate of
negative one generates a π¦-coordinate of zero. But there arenβt any other
π₯-coordinates to do that, so this is a curve which just touches the π₯-axis in one
place.

So letβs have a look at one more
example then. π¦ equals two π₯ squared plus π₯
plus three. So now π is two, π is one and π
is three. And plugging those numbers into our
quadratic formula gives us a discriminant of one squared minus four times two times
three; thatβs negative twenty-three. And when we try to solve this now,
weβve gotta find the square root of negative twenty-three. But you canβt get a square root of
a negative number because if I take a number multiplied by itself, whether itβs
positive or negative, Iβm always gonna get a positive answer.

So this is an example of a
quadratic thatβs got no roots. In other words, there arenβt any
π₯-coordinates which generate a π¦-coordinate of zero. We canβt find any real values of
π₯, which are gonna enable us to calculate this square root of a negative number
here. So we can use this information,
just by analyzing the discriminant, to tell whether there are gonna be two roots,
one root, or no roots for our quadratic equation. If π squared minus four ππ is
bigger than zero, then thereβre gonna be two roots. So you could just work out the
value of π squared, the value of four ππ, and if π squared is bigger than four
ππ, then you know that thereβre gonna be two roots. If π squared minus four ππ
equals zero, that square root is gonna be zero. So weβre just gonna have one
root.

And maybe a quicker way of spotting
that is, if that the π squared value is equal to four ππ, means the same thing;
thatβs just gonna be one root. And if π squared minus four ππ
is less than zero, youβre gonna be trying to calculate the square root of a negative
number. Itβs not gonna work; thereβll be no
roots. And that happens when the square of
π is less than four times π times π.

Okay then. Before we go, I just want you to do
these three questions.

How many roots do these quadratics
have? So weβre gonna be putting the
π¦-coordinate equal to zero and seeing how many solutions we get. And I want you to do this by
analyzing the discriminant in each case. So Iβm just- start- Iβve- a pause
now and Iβm just gonna wait a couple of seconds and then Iβm gonna explain the
answers.

Right. So in each case, the first thing to
do is write down the value of π, π and π. And then we can use those values to
evaluate the discriminant. And the discriminant, remember, is
π squared minus four ππ.

So with this first question, thatβs
five squared minus four times two times five. So thatβs twenty-five minus forty,
which is negative fifteen. So in this case, the discriminant
π squared minus four ππ was less than zero and that means that there are no real
roots.

Moving on to number two, we can see
that π is two, π is negative four, and π is two. So the discriminant is π squared
minus four ππ. So thatβs negative four squared
minus four times two times two. Well four squared is sixteen and
four times two is eight times two is sixteen. So weβve got sixteen minus sixteen,
so thatβs equal to zero. So π squared minus four ππ, the
discriminant, is equal to zero and that means weβve got one repeated root.

And for the last question, weβve
got π is equal to two, π is equal to one cause that means one times π₯, and π is
equal to negative three. And the discriminant π squared
minus four ππ is one squared minus four times two times negative three. Now four times two is eight times
three is twenty-four. So weβre taking away negative
twenty-four, so that means weβre adding twenty-four.

So watch out for these situations;
weβre taking away something but because one of those, π-value in this case, was
negative, weβve got the double negative thing. So the discriminant is twenty-five,
which is positive. So that means in the quadratic
formula, weβll be finding the square root of twenty-five, which would be positive or
negative five. So weβre adding or subtracting
something to our answer there. So the discriminant is greater than
zero, so in this case, number three, weβve got two roots.