### 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.