### Video Transcript

In this video, weβll learn how to
graph a quadratic function of the form π of π₯ equals ππ₯ squared plus ππ₯ plus
π in order to solve the equation π of π₯ equals zero.

Letβs begin by recalling what we
know about the graphs of quadratic functions. Theyβre the shape of a
parabola. And the coefficient of π₯ squared
tells us the orientation of this parabola. In particular, if the quotient of
π₯ squared that is π is greater than zero, we have that U-shaped parabola, and if
π is less than zero, we get an N-shaped parabola. In other words, if the coefficient
of π₯ squared is negative, we have the inverted parabola. In fact, we can also sketch such
graphs by calculating the values of their π₯- and π¦-intercepts. The values of their π₯- and
π¦-intercepts are found by setting π¦ equal to zero and π₯ equal to zero,
respectively, then either solving the resulting equation or simply simplifying.

Specifically for a quadratic
function of the form π of π₯ equals ππ₯ squared plus ππ₯ plus π, if the equation
ππ₯ squared plus ππ₯ plus π equals zero has two distinct solutions, π₯ equals π₯
one and π₯ equals π₯ two, then the intercepts are π₯ sub one and π₯ sub two as shown
here. Now this is really important when
it comes to using the graphs of such functions to solve equations. Since the π₯-intercepts of the
graph of a function π¦ equals π of π₯ are found by solving the equation π of π₯
equals zero, the reverse is true. It follows that the solutions to
the equation π of π₯ equals zero can be found by locating the π₯-intercepts of the
graph.

Now, of course, in the case of
quadratic graphes specifically, itβs a little more nuanced than this. Weβve just seen that if the graph
of a quadratic function has two distinct π₯-intercepts π₯ one and π₯ two, then the
equation π of π₯ equals zero has two distinct solutions. But there will be occasions where
our equation will have one solution, sometimes called a repeated root, or no
solutions at all. These can be quickly identified
again by looking at the graph of the function. A repeated root will occur when the
π₯-axis is a tangent to the graph. In other words, the graph touches
the π₯-axis exactly once.

On these occasions, in fact, the
single solution to the equation π of π₯ equals zero corresponds in fact to the
location of the vertex of the graph. Now, if the graph does not
intersect the π₯-axis at all, then the equation π of π₯ equals zero has no real
roots. This could look like this, in other
words, the function has purely positive outputs, or like this, where we have purely
negative outputs. So with all of this in mind, weβre
going to demonstrate how to use a quadratic graph to solve a quadratic equation.

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

Remember, given the graph of a
quadratic equation π of π₯ equals ππ₯ squared plus ππ₯ plus π, the solutions to
the quadratic equation π of π₯ equals zero correspond to the values of the
π₯-intercepts. So all we need to do is identify
the location of the π₯-intercepts on our graph. Locating these, we see that there
is, in fact, only one π₯-intercept. Itβs negative two. So thereβs actually only one
solution to the equation π of π₯ equals zero. Thatβs π₯ equals negative two. Now we have in fact been asked to
give the solution set to the equation π of π₯ equals zero. So in set notation, itβs the set
containing the single element negative two.

In our next example, weβll see how
we can perform the exact same process if the graph of the function is an inverted
parabola, in other words, if the coefficient of π₯ squared is negative.

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

Remember, if weβre given the graph
of a function π of π₯, the solutions to the equation π of π₯ equals zero
correspond to the values of the π₯-intercepts. And whilst weβre using this process
to find solutions to quadratic equations, this holds to equations of any function π
of π₯ equals zero.

So all we need to do is find the
location of the π₯-intercepts on our graph. We have two. Thereβs one here and thereβs one
here. These are the points of which our
graph passes through the π₯-axis. Since this occurs at negative two
and two, we can say that the solutions to the equation π of π₯ equals zero are π₯
equals negative two and π₯ equals two. Now in fact, we were asked to give
the solution set. So we use set notation as
shown. The solution set to the equation π
of π₯ equals zero is the set containing the elements negative two and two.

Weβll consider one further example
of this form.

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.

Given some graph of a function π¦
equals π of π₯, the solutions to the equation π of π₯ equals zero correspond to
the values of the π₯-intercepts if they exist. If there are no π₯-intercepts, it
follows that there are no solutions to the equation π of π₯ equals zero. Now in this case, weβve been given
the graph of the function π of π₯ equals π₯ squared minus two π₯ plus three and
asked to solve π of π₯ equals zero or, in other words, π₯ squared minus two π₯ plus
three equals zero.

But if we look carefully, we see
that there are no locations where the graph, which is the green plot here,
intersects the π₯-axis. And so, there are no real solutions
to the equation π of π₯ equals zero. In order to use set notation, we
need to find a way to show that there are no values inside the set. So we use the notation shown. Itβs the null set or empty set.

In the examples weβve been given so
far, weβve been given the graph of some function π of π₯. And that has allowed us to identify
the locations at which this graph crosses the π₯-axis. This information then allowed us to
write the roots of the function, in other words the solutions, in terms of the
associated solution set that could have either two distinct elements, one single
element, or no elements at all. And in the case if there only being
a single element, the location of the root is common with the only vertex of the
function. Now in fact, we wonβt always be
given the graph of the quadratic function. And it wonβt always be obvious how
to find exact solutions. In our next example, weβll plot the
graph of a quadratic function and use this to estimate the solutions to the
equation.

Plot the graph of the quadratic
function π of π₯ equals two π₯ squared plus three π₯ minus one over the interval π₯
equals negative three and π₯ equals one using integer values of π₯. Using this graph, estimate the
solutions to the equation π of π₯ equals zero. Give your answers to the nearest
whole number.

To plot the graph of a function, we
begin by drawing a table of values. Now, since weβre plotting it over
the interval π₯ equals negative three and π₯ equals one and using integer values,
weβre going to use π₯ equals negative three, negative two, negative one, zero, and
one. In order to find the corresponding
output, the value of π of π₯, we substitute each value of π₯ into the function two
π₯ squared plus three π₯ minus one. This creates a set of ordered pairs
that satisfy our function.

Letβs begin by calculating the
value of the function when π₯ equals negative three, in other words π of negative
three. We replace each instance of π₯ with
negative three, and we get two times negative three squared plus three times
negative three minus one. The order of operations then tells
us to calculate the exponent in this problem first. Negative three squared is nine. Then we calculate two times nine,
which is 18, and three times negative three, which is negative nine. So this becomes 18 minus nine minus
one, and that is equal to eight. So π of negative three is
eight.

Weβll now repeat this process with
π₯ equals negative two. Our expression is two times
negative two squared. Of course, negative two squared is
four. Then, we have three times negative
two minus one. That becomes eight minus six minus
one, which is one. Next, π₯ equals negative one. Itβs two times negative one squared
plus three times negative one minus one. And since negative one squared is
one, this becomes two minus three minus one, which is negative two. In a similar way, when π₯ is equal
to zero, the value of the function is negative one. And when π₯ is equal to one, π of
π₯ is equal to four.

And so we have our five ordered
pairs. We have a negative three, eight;
negative two, one; negative one, negative two; zero, negative one; and one,
four. We plot these in turn. Negative three, eight is here. Negative two, one is here. We then have negative one, negative
two; zero, negative one; and one, four.

Donβt forget, these outputs donβt
increase linearly. So we must join them with a smooth
curve instead of a straight line. And so, we have the graph of our
function. Remember, weβre trying to use this
to find solutions to the equation π of π₯ equals zero. Now, if these solutions exist, they
correspond to the values of the π₯-intercepts. These appear to be approximately
located at π₯ equals negative 1.8 and π₯ equals 0.2. Correct to the nearest whole
number, an estimate for solutions to the equation π of π₯ equals zero are π₯ equals
negative two and π₯ equals zero.

Now in fact, we donβt need to be
given or drawn the graph to find solutions to π of π₯ equals zero. We know that the solutions
correspond to the values of the π₯-intercepts, sometimes equivalently called the
zeros of the function. Hence, given these values or
coordinates of the π₯-intercepts, we can identify a solution set of the equation π
of π₯ equals zero.

In our next example, weβll
demonstrate what that looks like.

If the graph of the quadratic
function π cuts the π₯-axis at the points negative three, zero and negative nine,
zero, what is the solution set of π of π₯ equals zero in the set of real
numbers?

Now remember, if weβre given the
graph of a function, we can find the solutions to π of π₯ equals zero by locating
the π₯-intercepts or zeros of the function. Now, in this case, weβre not
actually given a graph, but we are told the coordinates at which the function cuts
the π₯-axis. Itβs negative three, zero and
negative nine, zero. Since the first number in each
ordered pair corresponds to the value of π₯ here, we can say that the solutions to
the equation π of π₯ equals zero are π₯ equals negative three and π₯ equals
negative nine. Using set notation, the solution
set of π of π₯ equals zero in the set of real numbers is the set containing the
elements negative three and negative nine.

Weβve now demonstrated in a variety
of ways how to find solutions to an equation π of π₯ equals zero when given the
graph of the function π of π₯. Letβs recap the key points from
this lesson.

We saw that the solutions to the
equation π of π₯ equals zero, if they exist, can be found by locating the values of
the π₯-intercepts of the graph of π¦ equals π of π₯. We also saw that if the π₯-axis is
a tangent to the curve, then we have a repeated root. Itβs just one root. And this point is also in fact the
vertex of the graph. Finally, we also saw that we can
plot quadratic graphs using a table of values. And we can then use this graph to
approximate the points at which the graph crosses the π₯-axis and hence the solution
or the roots of that quadratic equation.