### Video Transcript

In this video, we will learn how to
graph quadratic functions. We will consider those given in
standard and vertex form. Quadratic function graphs have many
applications. They can represent the paths of
moving objects like a bouncing ball or even flight patterns of bees. In business, they can forecast
revenue. We could also do something like
minimize waste and package sizing. When working with quadratic
function graphs, we often deal with variables such as speed, cost, and area, among
other things.

Before we look at any specific
graphs, letβs remind ourselves of the general form of a quadratic function. When we talk about quadratic
functions, the general form would be π of π₯ equals ππ₯ squared plus ππ₯ plus π,
where π, π, and π are constants and π cannot equal zero. Another form we might see quadratic
functions in looks like this. π of π₯ equals π times π₯ minus β
squared plus π, where π, β, and π are constants and π is not equal to zero. We call this form the vertex
form. Itβs called the vertex form because
when our quadratic function is written in this form, the vertex of this graph will
be at the point β, π.

If we have this sketch of a
quadratic function, the vertex would be found here. The vertex is the maximum in this
case because the graph opens downward. And when the quadratic opens
upward, the vertex is instead a minimum. We know that when π is greater
than zero, that is, when π is positive, the graph will open upward. And when π is less than zero, when
π is a negative value, the graph will open downward. We noted with the vertex form on
the right that when written in this form, the vertex is found at the point β,
π. When we consider the general form,
weβre not able to identify the vertex simply by looking at the general form.

However, we do have a formula to
find the vertex if we have a function in general form. From the general form, we take the
negative π-value and divide it by two times the π-value. This gives us the π₯-coordinate of
the vertex. We take that π₯-coordinate and we
plug it back into our function to solve for the π¦-coordinate of the vertex. Before we move on, we should
mention another vertex form. Some people in textbooks might use
the form π of π₯ equals π times π₯ plus π squared plus π, where π, π, and π
are constants and π does not equal zero.

This form operates very similarly
to the first vertex form we looked at, with one key distinction. If youβre using this form, the
vertex is found at negative π, π. By looking closely at these two
functions, we can see why. In the first form, weβre
subtracting β from π₯. And then we have a positive vertex
π₯-coordinate. In the second form, we are adding
that constant to π₯, and weβll therefore need to take the negative value as the
π₯-coordinate for the vertex. Both forms are equally acceptable
as long as youβre using them consistently.

To prepare to graph a quadratic
function, weβll consider four different features that will help us graph these
functions. First, weβll consider the shape of
the parabola. Second, weβll consider the
π¦-intercept of quadratic functions. Third, weβll consider the roots or
zeros of the function, which are the π₯-intercepts of the function. And finally, weβll consider the
vertex or turning point of the function. To help us consider each of these
features, weβll look at the function π of π₯ equals π₯ squared minus two π₯ minus
eight. To find the shape of the parabola,
weβll want to know is the π-value greater than zero or less than zero? That is, is the π-value positive
or negative?

Positive π-values open upward, and
negative π-values open downward. In our case, the coefficient of the
π₯ squared term is one. One is greater than zero. Therefore, the shape of this graph
will open upward. Next, we want to find the
π¦-intercept. The π¦-intercept will be the place
when π₯ equals zero. For us, that is zero squared minus
two times zero minus eight. π of zero equals negative eight,
which makes the π¦-intercept zero, negative eight. Itβs worth pointing out here that
when weβre dealing with the general form of a quadratic function, the π¦-intercept
is always found at the point zero, π. In our function, π is equal to
negative eight, which confirms a π¦-intercept of zero, negative eight.

What about the roots of a quadratic
function? The roots are the place when the π
of π₯ equals zero. To find the roots, we set our π of
π₯ equal to zero and then solve for π₯. When the equation is given in
general form, we can solve by factoring. We have two π₯-terms. We need factors that when
multiplied together equal negative eight and when added together equal negative two,
which will be negative four and positive two. We then need to take each of these
factors and set them equal to zero. We need to know when π₯ plus two
would be equal to zero and when π₯ minus four would be equal to zero. That would be when π₯ equals
negative two and when π₯ equals positive four. This is telling us that this
quadratic function graph will cross the π₯-axis at the point negative two, zero and
the point four, zero.

Our final feature, the vertex,
would be given in the form β, π if our original function was written in vertex
form. But since we only have the general
form, weβll need to calculate negative π over two π for the π₯-coordinate and then
plug that in to solve for the π¦-coordinate of this vertex. Negative π over two π would be
the negative of negative two, so positive two over two times one. Positive two over two equals
one. This means that the π₯-coordinate
of our vertex is one. To find the π¦-coordinate of the
vertex, we plug in one for π₯. We get one squared minus two times
one minus eight, which equals negative nine. And we can say the vertex of this
function is located at one, negative nine. And because we know the shape of
this function, we can say that the vertex will be a minimum coordinate.

And with these key features, weβre
ready to sketch a graph. We know our π¦-intercept is located
at zero, negative eight. We have roots at negative two, zero
and four, zero. And our vertex, our minimum
coordinate, is at one, negative nine. You might also remember that these
graphs are symmetric about the axis of symmetry, which is located along the line of
the π₯-coordinate of the vertex. Thinking about the axis of symmetry
gives us a better idea of how to sketch the graph. And weβll do that now by connecting
these points. And following all these features,
weβve graphed a quadratic function. Using these four features, weβre
ready to consider some examples.

Write the quadratic equation
represented by the graph shown.

In order to write this equation,
letβs consider some features of the graph of quadratic functions, shape,
π¦-intercept, the roots, and the vertex. We also have the general form π of
π₯ equals ππ₯ squared plus ππ₯ plus π or the vertex form π of π₯ equals π times
π₯ minus β squared plus π. Considering the shape, we see that
this parabola opens downward. And that means the π-value will be
less than zero. It will be negative. The π¦-intercept is located here at
zero, zero, which mean that the roots here will just be one root at zero, zero. And it happens to be the vertex,
which is the maximum point of this function.

If we use vertex form, we know that
the vertex is found at the point β, π. And so we can take this vertex of
zero, zero and plug it into that general vertex form. When we simplify that, we find out
that the function is π times π₯ squared. And that means we need to know what
π is. We know that the π-value is
negative. But in order to find what it
exactly is, we need to consider another point from the graph. We could use one of the other
points we know from the graph. For example, we know that the graph
crosses the point two, negative four. So we plug in two for π₯ and
negative four for π of π₯. And then we have negative four
equals four π.

From there, we divide both sides of
the equation by four. And we see that negative one equals
π or π equals negative one. And we plug that value back in for
π, which gives us negative one times π₯ squared. And we can simplify that to just be
negative π₯ squared. So we found the quadratic equation
represented by the graph shown to be π of π₯ equals negative π₯ squared.

In our next example, we start with
the quadratic equation and we need to find its graph.

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

Our function π of π₯ equals
negative two π₯ squared plus nine π₯ minus seven is given in the general form ππ₯
squared ππ₯ plus π. But before we do anything with the
function, letβs consider each of these graphs. When we look at the graph of (A),
it opens upward and has a vertex in the fourth quadrant. The graph of (B) opens downward and
has a vertex in the second quadrant. (C) opens downward and has a vertex
in the first quadrant. (D) opens downward and has a vertex
in the first quadrant. And (E) opens upward and has a
vertex in the fourth quadrant.

A good strategy to finding the
graph would be to take our equation and find its vertex. When we have an equation in the
general form, we find the vertex to be negative π over two π. Thatβs the π₯-coordinate. So we take negative nine over two
times negative two, which is positive 2.25. This means the π₯-coordinate of our
vertex will be located at 2.25. If we draw the line π₯ equals 2.25
on all five of our graphs, we see the in (A) that intersects the vertex and in (C)
that intersects the vertex. On that basis, (B), (D), and (E)
are not the graphs weβre looking for.

To find the π¦-coordinate of the
vertex, we could plug in 2.25 into our equation. When we solve for that, we get
3.125. And the vertex for (C) is located
at 3.125. We also notice that our π-value is
negative. And that means our graph must open
downward. Option (A) opens upward and has a
vertex in the wrong place. And so we can say that option (C)
is the correct graph here.

Hereβs another example where weβll
use a graph to write a quadratic equation.

Write the quadratic equation
represented by the graph shown.

We have the general form for a
quadratic equation π of π₯ equals ππ₯ squared plus ππ₯ plus π and the vertex
form π times π₯ minus β squared plus π. To find the equation, weβll
consider some features of the graph. The shape of this graph opens
downward. And that means we know that our
π-value will be less than zero. It will be negative. We have a π¦-intercept at the point
zero, two. When it comes to the roots, we
canβt know with a great deal of accuracy what the roots are. So weβll leave them there for
now. The vertex here is a maximum. And we do know where itβs located,
at the point one, three.

Because we know the vertex, we can
start with our vertex form. Our vertex is β, π. And so we plug in one for β and
three for π. The only thing weβre missing now is
this π-variable. We know that itβs less than zero,
but we donβt know exactly what it is. To find it, we can plug in another
point from the graph that we already know. If we plug in zero for π₯ and two
for π of π₯, weβll be able to solve for our π-value. Zero minus one squared is one; one
times π is π, which means π plus three equals two. And two minus three is negative
one, so we can say that π equals negative one.

And weβll go back and plug that
in. Instead of having negative one, we
can just write the negative sign and say that π of π₯ equals negative π₯ minus one
squared plus three. This is the vertex form of the
graph we have. If we wanted to write this in the
general form, we could expand this π₯ minus one squared, which would give us the
negative of π₯ squared minus two π₯ plus one plus three. We distribute the negative. And when we combine like terms, we
have the general form of negative π₯ squared plus two π₯ plus two. Both of these forms are the
quadratic represented by this graph.

In our final example, weβre not
looking for a vertex form or a standard form of a quadratic function. Instead, weβre going to explore the
factored form.

Write the quadratic equation
represented by the graph shown. Give your answer in factored
form.

To do this, letβs consider some
features of the graph of quadratic functions like shape, π¦-intercept, roots, and
vertex. The shape of this parabola opens
upward. And that tells us that our π-value
will be positive, greater than zero. We have a π¦-intercept located at
the point zero, zero. To find the roots, we look for the
place where the graph crosses the π₯-axis. Here we have a root at zero, zero
and at five, zero. The vertex of this graph has been
labeled. Itβs a minimum coordinate located
at two and a half, negative 6.25.

What we wanna do now is think a
little bit more about the roots. The roots are sometimes called
solutions. Theyβre the place where our
function equals zero. We have solutions when π₯ equals
zero and when π₯ equals five. We can take these solutions and
turn them into factors. Because π₯ is the same thing as π₯
minus zero, π₯ equals zero is a factor. And for factor two, we would want
to say π₯ minus five equals zero. Here are two factors. However, we still donβt know what
our π-value is because there are many equations that have the factors π₯ and π₯
minus five. And that means we should have π of
π₯ equals π times π₯ times π₯ minus five.

To solve for π, we can plug in
coordinates of points we know fall on this graph. We know that when π₯ equals 2.5, π
of π₯ equals negative 6.25. Two and a half minus five is
negative two and a half. Two and a half times negative two
and a half equals negative 6.25. And if we divide both sides of the
equation by negative 6.25, we see that π equals one. And if π equals one, then our
factored form is π of π₯ equals π₯ times π₯ minus five. We always want to check for this
π-value because as Iβve just drawn on top of this graph β hereβs an equation that
has the same factors. However, in this case, the π would
have to be less than zero because the graph opens downward, which means itβs not
enough just to know the factors. You need to also check for the
π-value.

Before we finish, weβll summarize
what weβve learned. Using the forms of quadratic
functions, π of π₯ equals ππ₯ squared plus ππ₯ plus π and π of π₯ equals π
times π₯ minus β squared plus π, we can identify features of the graph. For the general form π of π₯
equals ππ₯ squared plus ππ₯ plus π, when π is greater than zero, the parabola
opens upward. When π is less than zero, the
parable opens downward. The π¦-intercept is located at
zero, π.

To find the π₯-coordinate of the
vertex, you take negative π over two π. To find the π¦-coordinate of the
vertex, you find the π₯-coordinate and then plug that back into your function. And for π of π₯ equals π times π₯
minus β squared plus π, the vertex form. When π is greater than zero, the
parabola opens upward. When π is less than zero, the
parabola opens downward. And the vertex is located at the
point β, π.