### Video Transcript

In this video, we will learn how to
solve quadratic equations using function graphs. First, we remember that quadratic
equations could be in the form π¦ equals ππ₯ squared plus ππ₯ plus π, where π,
π, and π are constants and π cannot equal zero. This is the standard form for
quadratic equations. We might also see them as π¦ equals
π times π₯ minus β squared plus π, where π is a constant and β, π is the vertex
or the turning point of the function. And we call this the vertex
form. But weβre not just looking at
quadratic equations. Weβre looking at how to solve
them.

Solving an equation is to find the
values of π₯ for which π¦ is zero. One of the ways we solve quadratic
equations is by factoring. For example, if we wanted to solve
π¦ equals π₯ squared plus four π₯ plus three by factoring, we set π¦ equal to
zero. And then we take out the terms π₯
plus one and π₯ plus three. We set both factors equal to
zero. And then we get the values four π₯
which makes π¦ equal to zero in this equation. And we call these roots or
solutions. But in this video, we want to solve
by graphing. Weβll do this by identifying points
of interest on the graph. To do that, weβll consider the
characteristics of quadratic graphs.

Some important characteristics of a
quadratic graph. Quadratic-function graphs have a
distinctive parabola shape. We generally think of it as a U
shape pointing upward or downward. They will open upward when π is
greater than zero, when π is positive. Remember that π-value is the
coefficient of the π₯ squared term when we write our quadratic equation in standard
form. And likewise, the graph will open
downward when π is less than zero, when π is negative. We can also say that when π is
positive, there will be a minimum value on the graph and that when π is negative,
there will be a maximum value on the graph.

And we remember that π cannot be
equal to zero as that would mean the graph would not be quadratic but linear. We also know that
quadratic-function graphs are symmetrical about the vertex. We could fold the graph in half at
the vertex. And both sides would lie on top of
each other. The quadratic-function graphs have
a π¦-intercept at the point zero, π. The graphs will have π₯-intercepts
or π₯-intercept where π¦ equals zero. Remember, these are also called
roots or solutions. The graph could have two solutions,
one solution, or no solutions if the graph does not cross the π₯-axis at all.

To solve a quadratic function by
graphing, we will be trying to identify the places where the graph crosses the
π₯-axis. Letβs try to do that now.

The diagram shows the graph of π¦
equals π of π₯. What is the solution set of the
equation π of π₯ equals zero?

We know that π¦ equals π of
π₯. And weβre looking for the place
where π of π₯ equals zero. That will be the place for this
quadratic where π¦ equals zero. And π¦ equals zero at the
π₯-intercept. π¦ equals zero along the
π₯-axis. And this function is crossing the
π₯-axis at negative two. Therefore, the π₯-intercept is
negative two, zero. But the solution set will be the
π₯-value that makes π of π₯ equal to zero. And that means instead of
coordinate form, weβre only interested in the π₯-coordinate. This function equals zero when π₯
equals negative two. Therefore, the solution set will
only include the value negative two.

Hereβs another example.

Consider the graph. The roots of the quadratic can be
read from the graph. What are they?

We should immediately be thinking
about what the roots of a graph are. The roots of a quadratic equation
are the π₯-values that make π¦ equal to zero. And that means we need to look at
the places this graph crosses the π₯-axis. The π₯-axis is the line π¦ equals
zero. And here, we see the two places our
graph is crossing.

Between zero and negative one,
there are five spaces. And our point falls in the middle
of the third space. It falls halfway between zero and
negative one. That first location is then π₯
equals negative one-half. When π₯ equals negative one-half,
π¦ equals zero. And the second root is halfway
between negative one and negative two, which is negative one and a half. But for consistency, we could write
it as negative three-halves. And so we can say that the roots of
this graph are negative one-half and negative three-halves.

Hereβs a third example.

The graph shows the function π of
π₯ equals π₯ squared minus two π₯ plus three. What is the solution set of π of
π₯ equals zero?

If weβre looking for π of π₯ equal
to zero, weβre looking for the π₯-values that make π₯ squared minus two π₯ plus
three equal to zero. And on the graph, that will be the
place where π¦ equals zero. And we know that π¦ equals zero
along the π₯-axis. In order for this function to have
a solution of π of π₯ equals zero, it would need to cross the π₯-axis. This function does not cross the
π₯-axis. Therefore, it has zero solutions or
zero roots. And therefore, the solution set
here is the empty set; there is no solution.

In our next example, weβll draw our
own graph to solve an equation.

By drawing a graph of the function
π of π₯ equals two π₯ squared minus three π₯, find the solution set of π of π₯
equals zero.

Our equation is π of π₯ equals two
π₯ squared minus three π₯. Before we start drawing our graph,
itβs helpful to have a few coordinates so that we can sketch the curve. We can use the tables to do
this. If we calculate the π₯-values,
negative two, negative one, zero, one, two, that should give us some idea of the
shape of this graph. Our first box would be two times
negative two squared minus three times negative two, which equals 14. Next, we have two times negative
one squared minus three times negative one, which is five. When we plug in zero, the result is
zero. When we plug in one, we get
negative one. When we plug in two, we get
two.

The top values, the π₯-values,
represent the domain of the function, what we can plug in for π₯. And the bottom values represent the
range, what weβll need for the π¦-values. In these values, we have a range
thatβs lowest point is negative one and highest point is 14. Now, these are just from the points
weβve chosen. That doesnβt mean itβs the range of
the full function. But it does give us an idea of what
the scale of the π₯- and π¦-axis should be. We can graph a point at negative
two, 14; negative one, five; zero, zero; one, negative one; and two, two.

When weβre looking at this, we see
that it might be helpful to have an additional point on the π₯-axis, so we could
solve for three. When we plug three into the
equation, we get nine as the output. And that just gives us one more
point to be able to sketch this graph. It wonβt be perfect, but you can
try to get a smooth curve between the points. And when we do that, since weβre
looking for the solution set of π of π₯ equals zero, weβre looking for the
π₯-intercepts. Weβre looking for the places where
this function crosses the π₯-axis. The first one is very clear. Itβs the point zero, zero. And the second intersection happens
halfway between one and two. We have a solution when π₯ equals
zero and a solution when π₯ is one and a half which makes our solution set zero and
three-halves or zero and one and a half.

Remember, this is not a coordinate
point. These are two different π₯-values
which when inputted into the function, the output is zero.

Now, weβre going to look at an
example where weβll use the solutions of the equation, the roots, to help us draw a
graph. This can be a quicker method than
using a table of values every time. However, weβll still need to
identify some other coordinates to help us draw a graph.

(1) Solve π₯ squared minus 10π₯
plus 25 equals zero by factoring. (2) Draw the graph of π of π₯
equals π₯ squared minus 10π₯ plus 25.

For the first part of this problem,
weβll need to factor π₯ squared minus 10π₯ plus 25. Since π₯ squared has a coefficient
of one, we know that each of the factors will have one π₯ term. And then we need to consider two
values that when multiplied together equal positive 25, but when added together
equal negative 10. We have one and 25; negative one,
negative 25; positive five, positive five; and negative five, negative five.

When we add negative five plus
negative five, we get negative 10. And so we found the pairs. And we realize that this is π₯
minus five times π₯ minus five. Or we could say itβs π₯ minus five
squared. But again We need to solve for π₯
here. And that means we should get π₯ by
itself, take the square root of both sides, and we get π₯ minus five equals
zero. And so we can say that the solution
or the root of this equation is π₯ equal to five. The equation only has one root. And that means it will only cross
the π₯-axis one time. This is all we need for the first
part of the question.

But in order to draw this graph,
weβll need a little bit more information. We know we have the root five,
zero. And because there is only one root,
because this is a repeated root, the point five, zero will also be the vertex of the
equation. Our equation was originally given
to us in standard form π¦ equals ππ₯ squared plus ππ₯ plus π. And when the function is given in
standard form, the π¦-intercept is located at zero, π. For us, that constant value, that
π-value, is 25. And we can say that we have a
π¦-intercept at zero, 25.

We can use these values to go ahead
and sketch our π₯- and π¦-axis. Because the coefficient of our π₯
squared, our π-value, is positive one, we can say that the parabola will be opening
upward and we can say that the vertex will be a minimum. Since five, zero is the minimum
value, we can say that zero is the lowest value for our π¦-coordinates. Then weβll add the scale to our
graph using this information. We have a minimum at five, zero, a
π¦-intercept at zero, 25. And because we know that parabolas
are symmetrical about the vertex, the 25 is five units away from the π₯-coordinate
of the vertex. And that means five coordinates to
the right of the vertex would also be equal to 25. There would be a coordinate at 10,
25 as well as at zero, 25. And from there, we try and use a
smooth curve to connect the points, which completes our graph.

In all of our previous examples,
weβve been solving for the π₯-values when π¦ equals zero. Weβll now look at an example where
we can use the intersection of a quadratic curve and another line.

The graph shows the function π of
π₯ equals two π₯ squared minus four π₯ minus six. What is the solution set of π of
π₯ equals zero? What is the solution set of π of
π₯ equals negative six?

The solution set for π of π₯
equals zero will be the places where two π₯ squared minus four π₯ minus six equals
zero. And on the graph, thatβs the place
where they cross the π₯-axis. The π¦ equals zero across the
π₯-axis. In this case, we have a solution at
negative one and at positive three. So for π of π₯ equal to zero, the
solution set is negative one, three. Remember, this is not a
coordinate. These are the π₯-values for which
π of π₯ equals zero. And weβve given them in set
notation.

The π₯-axis, as weβve said, is the
place where π¦ equals zero. The line π of π₯ equals zero is
the π₯-axis. But what would we do if we were
using a graph to look at π of π₯ equals negative six? This would be the place or the
places where our quadratic cross the line π¦ equals negative six. And in this graph, we see thatβs
happening twice. It happens when π₯ equals zero and
when π₯ equals two. In set notation, weβre saying the
two π₯-values for which π of π₯ equals negative six are zero and two.

Letβs review the key points when
dealing with solving quadratic equations graphically. Solving quadratic equations
graphically works by inspecting the graph of a function. This process yields an approximate
solution. Therefore, when dealing with
noninteger roots, it may be better to solve for the roots algebraically. We remember that quadratic
equations can have one, two, or zero roots. And when we look at this on the
graph, that means it can cross the π₯-axis one time, two times, or not at all.

And finally, to solve a quadratic
equation graphically, we find the place on the graph where the equation crosses the
π₯-axis. It might be worth noting here that
this can be a useful method if youβre using a graphing calculator to solve this kind
of problem. You can enter quadratic equations
into your graphing calculator. And it will produce a graph for you
to use. Even when working with a
calculator, the procedure for identifying the π₯-intercepts will remain the same,
looking for the places where the equation crosses the π₯-axis.