### Video Transcript

In this lesson, weβll learn how to
graph any quadratic function using a table of values and a given interval and
identify the features of the graph.

Now, quadratic equations are used
in everyday life. Theyβre used in science, business,
and engineering. For instance, they can determine
the paths of moving objects, from bouncing balls to the flight path of a
projectile. Businesses use them to forecast
revenues and design packaging to minimize waste. We can also use quadratic equations
to identify the minimum and maximum values of many different variables, such as
speed, cost, and area. In particular, we can use the
graphs of quadratic functions to easily determine information about maximum and
minimum values and when the outputs will be equal to zero.

To do this, letβs begin by
reminding ourselves what we mean when we say a function is quadratic. This is a single-variable
polynomial function of degree two. In other words, a quadratic
function has the form π of π₯ equals ππ₯ squared plus ππ₯ plus π for constants
π, π, and π, where π cannot be equal to zero. Then, for this quadratic function
π of π₯, π¦ equals π of π₯ is a quadratic equation.

Letβs now consider the graphs of
two quadratic equations. Weβll begin by looking at π¦ equals
π₯ squared plus four and a second quadratic equation π¦ equals negative π₯ squared
plus nine. The graphs look a little something
like this. And we might notice that they have
a very similar shape. This shape is known as a
parabola. In fact, all quadratics have this
shape. And the sign of the constant value
π, the coefficient of π₯ squared, tells us whether the parabola opens upwards or
downwards. In particular, if the coefficient
of π₯ squared is positive, if π is greater than zero, the parabola opens
upward. And if π is less than zero, the
parabola opens downward.

Letβs identify some of the other
useful points on our graph, firstly the π¦-intercept, the point on each graph where
it passes through the π¦-axis. In these cases, the π¦-intercepts
are four and nine, respectively. But of course we might recall that
we can find the value of the π¦-intercept by setting π₯ equal to zero and
substituting that into each of our equations. But letβs look at what happens if
we substitute π₯ equals zero into the general form of our quadratic function. We get π times zero squared plus
π times zero plus π, which is simply equal to π. So, in fact, the value of the
π¦-intercept of a quadratic function in this form is simply π. Or we can say it has coordinates
zero, π.

In a similar way, we can in fact
find the value of the π₯-intercepts if they exist by substituting π¦ equals zero or
π of π₯ equal to zero and solving for π₯. Letβs demonstrate this in our first
example.

Identify the π₯-intercepts of
the quadratic function π of π₯ equals π₯ squared minus one.

And then we have a graph of
that function shown. Now, in fact, there are two
ways to answer this question then. We could simply look at the
coordinates of the points where this graph passes through the π₯-axis. In fact, weβll use that
technique to check our answer. Instead, we recall that the
π₯-intercepts occur when π of π₯ or π¦ is set equal to zero. In this case then, we need to
set the expression π₯ squared minus one equal to zero and solve for π₯. We could factor the expression
on the left-hand side, or we can simply add one to both sides to get π₯ squared
equals one. And then we can take the square
root of both sides of this equation, remembering that in fact weβll need to take
both the positive and negative square root of one. So π₯ is equal to positive or
negative root one. But of course the square root
of one is simply one. So π₯ is either equal to one or
negative one.

We can denote these using
subscript and work in ascending order. So π₯ sub one equals negative
one and π₯ sub two equals one. Letβs check these on the
graph. We begin by identifying the
points at which the graph passes through the π₯-axis, as shown. Then, we look at the scale. In fact, we observe that four
small squares is equivalent to two units. This means that two small
squares must be equal to one unit. So these points are indeed
negative one and one. The π₯-intercepts of the
quadratic function given are π₯ sub one equals negative one and π₯ sub two
equals one.

Now, before we move on, letβs
inspect this graph a little more. We first notice that the turning
point is in fact the minimum point. Itβs the point on the curve which
outputs the smallest possible value of the function. We also see that if we construct a
vertical line through this point, through the turning point of the graph, that line
is the line of symmetry. And of course this point has a
special name. Itβs called the vertex. And the vertex is really useful
because it tells us the value of the minimum or maximum output of the function
depending on the sign of the coefficient of π₯ squared. And we can note that since
parabolas are symmetric in the vertical line through their axis of symmetry, this
line must in fact lie halfway between the π₯-intercepts, if they exist. In fact, it will lie halfway
between any π₯-values as shown on the graph with the same output.

Letβs summarize all of the
properties of quadratic functions and their graphs. Remember, for the function π of π₯
equals ππ₯ squared plus ππ₯ plus π, if π is positive the graph opens upward and
if itβs negative it opens downward. The π¦-intercept, which could be
found by setting π₯ equal to zero, is actually just given by the point zero, π. And that can be zero, one, or two
π₯-intercepts.

Now, these π₯-intercepts can be
found if they exist by solving the equation π of π₯ equals zero. The single turning point of the
function is called its vertex, and it will either be a maximum or a minimum. And of course these parabolas are
symmetric in the vertical line through their vertex. We also saw that the axis of
symmetry lies halfway between the π₯-intercepts, if they exist.

In our next example, weβre going to
identify the axis of symmetry for a given quadratic curve.

Identify the axis of symmetry
of the quadratic function π of π₯ equals π₯ squared minus one.

Remember, when we talk about
the axis of symmetry of a quadratic curve, weβre talking about the vertical line
that passes through its vertex or turning point. Now, whilst we could use the
diagram to do so, weβre going to use an extra fact here. That is, the axis of symmetry
will actually lie exactly halfway between the π₯-intercepts of the graph. In fact, it will lie halfway
between any π₯-values which have the same output.

Letβs set π of π₯ equal to
zero in order to find the value of the π₯-intercepts and then double-check this
by looking at the graph. When we do, we get π₯ squared
minus one equals zero. We add one to both sides and
then take the positive and negative square root of one. So π₯ is positive or negative
root one, and the two π₯-intercepts are negative one and one. And we can see they do indeed
match the values on our graph.

We said that the axis of
symmetry will actually lie exactly halfway between these two values. And of course we find their
midpoint by finding their mean average. We add them together and divide
by two. Thatβs negative one plus one,
which is zero, divided by two, which is still zero. And so the axis of symmetry
will pass through the point π₯ equals zero, and itβs a vertical line. Hence, the axis of symmetry has
the equation π₯ equals zero.

Inspecting the graph, this does
indeed look likely. But letβs check by choosing
values of π₯ which have the same output. The point exactly halfway
between negative four and four is zero, confirming to us that the axis of
symmetry of our quadratic function is π₯ equals zero.

Now, up to this point, weβve used
graphs and their equations to identify various features. In our next example, weβll
demonstrate how we can actually use a table of values to help us sketch the graph of
a quadratic function.

Which of the following graphs
represents the equation π of π₯ equals negative two π₯ squared plus nine π₯
minus seven?

There are two ways we can
answer this question. But the first thing that weβre
going to do is establish whether the graph of our function opens upwards or
downward. Firstly, we know that if we
have the equation π of π₯ equals ππ₯ squared plus ππ₯ plus π and the value
of π is positive, the graph opens upward. If itβs negative, it opens
downward.

Now, for our function, the
value of π is negative two. So itβs negative. This means the function opens
downward. And we can therefore disregard
any functions that open upward. Thatβs the function represented
by graph (B) and the one represented by graph (E).

Now that weβve done this, weβre
going to plot a table of values to identify the points through which this graph
passes. Weβll need to do this for a
variety of values of π₯. Letβs start with π₯ equals
zero. Thatβs π of zero equals
negative two times zero squared plus nine times zero minus seven. And thatβs negative seven. So when π₯ is equal to zero,
the value of the function is negative seven. And that actually means that
the π¦-intercept of our function is negative seven.

But of course the other
technique we couldβve used is to identify that for the function ππ₯ squared
plus ππ₯ plus π, the π¦-intercept has coordinates zero, π. Either technique here,
substituting or identifying the feature of the graph, is equally acceptable.

Weβre now going to choose some
positive and negative values of π₯ to satisfy each of the graphs weβve been
given. First, letβs substitute π₯
equals negative three into the function. We get negative two times
negative three squared plus nine times negative three minus seven. Thatβs negative 52. Similarly, letβs substitute π₯
equals negative two. When we do, we find that π of
negative two is negative 33. And that means our graph must
pass through the point negative two, negative 33. Substituting the remaining
values, and we see that π of negative one is negative 18, π of one is zero, π
of two is three, and π of three is two. Plotting the values that we can
on our graphs, and we see we disregard option (C) completely. We notice itβs the wrong side
of the π¦-axis. And in fact the graph that does
pass through these points is graph (A).

Now, in fact, there was an
alternative method here. We began by identifying the
shape of the graph. We saw that it opened
downward. And we found the value of its
π¦-intercept. We could also have set π of π₯
equal to zero to find the value of the π₯-intercepts. Doing so wouldβve given us π₯
equals one and π₯ equals 3.5, which once again matches function (A). So the graph that represents
the equation given is graph (A).

Letβs finish by recapping some of
the key points from this lesson. In this lesson, we learned that a
quadratic function is of the form ππ₯ squared plus ππ₯ plus π, where π is not
equal to zero. We learnt that all quadratics are
in the shape of a parabola, and they open upwards when π is positive and down when
π is negative. And we learned that whilst we could
set π₯ equal to zero to find the value of the π¦-intercept for the equation π¦
equals ππ₯ squared plus ππ₯ plus π, the π¦-intercept actually has coordinates
zero, π.

We learned that the value of the
π₯-intercepts, if they exist, can be found by solving π of π₯ equals zero. We learned also that graphs of
quadratics have a single point called the vertex, and the sign tells us whether this
will be a maxima or a minima. This vertex lies halfway between
the π₯-intercepts. And of course quadratics are
symmetric in the vertical line that passes through this vertex, which we call the
axis of symmetry.