Video Transcript
In this lesson, we’ll learn how to
graph any quadratic function that’s given in its standard and vertex forms using and
studying its transformations.
Quadratic equations are used in
everyday life. They’re used in science, business,
and engineering. They can help us to model paths of
moving objects, from bouncing balls to the flight paths of bees. Businesses use them to forecast
revenues and design packaging to minimize waste. Further, we can use quadratic
equations to identify minimum and maximum values of so many different variables,
including speed, cost, and area.
So let’s begin by recalling what we
mean when we say that an equation is a quadratic. A quadratic equation is one that
can be written in the form 𝑦 equals 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐. Now it’s important that 𝑎 is not
equal to zero and 𝑎, 𝑏, and 𝑐 are all real constants. In fact, the word “quadratic” comes
from the Latin for square. These are all equations where the
highest power of 𝑥 is two; it’s squared. So how do we sketch these
graphs? Let’s begin by reminding ourselves
what the graph of 𝑦 equals 𝑥 squared looks like. It has that typical parabola
shape. It passes through the origin. And this is, in fact, the location
of its vertex or turning point.
The graph of 𝑦 equals negative 𝑥
squared is found by reflecting this graph in the 𝑥-axis. So it’s an inverted parabola with
its vertex at the exact same point. But what about if we apply further
transformations? For instance, imagine we wanted to
draw the graph of 𝑦 equals 𝑥 plus four squared. In fact, this is a horizontal
translation. The graph is going to move four
units left. And so we see the vertex of this
graph lies at negative four, zero. Then what about the graph of 𝑦
equals 𝑥 plus four squared plus two? Well, this time, that’s a vertical
translation of our previous graph of 𝑦 equals 𝑥 plus four squared. It’s going to move two units
up. And so it will have a vertex at
negative four, two.
So how does this help? We don’t want to have to apply a
series of transformations every time we sketch a graph. Well, it means if we can rewrite
our quadratic equation in the form 𝑎 times 𝑥 plus 𝑘 all squared plus ℎ, where 𝑘
and ℎ are real numbers, we can say that this graph has a vertex at the point
negative 𝑘, ℎ. This, of course, is in completed
square form. So combining this with what we know
about sketching any graph, and we have a handy how to guide, we begin as before. We simply check the shape of the
parabola. If 𝑎 is greater than zero — in
other words, if the coefficient of 𝑥 squared is positive — we have the standard
parabola shape, where its vertex is a minimum. It’s the lowest point on the
graph.
If, however, 𝑎 is less than zero —
in other words, the coefficient of 𝑥 squared is negative — we have an inverted
parabola. And so the vertex is in fact a
maximum point. Then just as we would if we were
sketching the graph of a linear function, we find the 𝑦-intercept by setting 𝑥
equal to zero and solving for 𝑦. Similarly, we can find the location
of any 𝑥-intercepts by setting 𝑦 equal to zero. Of course, with these graphs, it’s
very possible that there are no 𝑥-intercepts at all. In this case, setting 𝑦 equal to
zero and solving for 𝑥 would result in no real solutions. We can find the location of the
vertex by writing in completed square form. So 𝑎 brackets 𝑥 plus 𝑘 all
squared plus ℎ has a vertex of negative 𝑘, ℎ. And so, with these four steps, we
can then sketch or even recognize graphs of quadratic functions. Let’s demonstrate this in our first
example.
Which graph represents the function
𝑦 equals negative 0.5𝑥 squared plus four?
Here, we have a quadratic
equation. We can identify its graph by
performing a series of steps. We begin by simply identifying the
correct shape. We know that if the coefficient of
𝑥 squared is positive, then we have the standard parabola. But if the coefficient of 𝑥
squared is negative, we have an inverted parabola. In this case, 𝑎, the coefficient
of 𝑥 squared, is negative 0.5. And that’s less than zero. So we have an inverted
parabola. This means we cannot choose (a) or
(b) because they have the standard parabola shape. Next, we can find the location of
the 𝑦-intercept by setting 𝑥 equal to zero. When we do, we find that the
equation becomes 𝑦 equals negative 0.5 times zero squared plus four. And that’s equal to four. So we know that the 𝑦-intercept is
at zero, four. This means we can very quickly
disregard option (d) also; that has a 𝑦-intercept at negative four.
And so that really only leaves us
with one option, option (c). We’ll check this by letting 𝑦
equals zero and solving for 𝑥. And this will tell us the location
of any 𝑥-intercepts. That’s zero equals negative 0.5𝑥
squared plus four. Adding 0.5𝑥 squared to both sides
gives us 0.5𝑥 squared equals four. And then when we divide through by
0.5, we get 𝑥 squared is equal to eight. We can then take the square root of
both sides of this equation. And we must remember to take both
the positive and negative square root of eight. So the 𝑥-intercepts or the roots
of this equation are positive and negative root eight.
Then we can estimate the value of
root eight by recognizing that it’s between the square root of four and the square
root of nine. So, in fact, it’s between two and
three. And since eight is closer to nine
than it is to four, the solution is likely to be closer to three than it is to
two. Well, we can see the 𝑥-intercepts
on our diagram are a little bit greater than negative three and a little bit less
than three. And so the graph that represents
the function given is (c).
In our next example, we’ll look at
how to manipulate a quadratic equation to find its graph.
Which of the following graphs
represents the equation 𝑦 equals 𝑥 squared minus five 𝑥 plus eight?
This is a quadratic equation. So there are a few things we can do
to help us identify its graph. First, for a quadratic equation of
the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐, if the coefficient of 𝑥 squared 𝑎 is
greater than zero, then it’s the usual parabola shape we expect. But if 𝑎 is less than zero, then
it’s an inverted parabola. Well, in our case, 𝑎 is simply
equal to one. We have one 𝑥 squared. So it’s greater than zero. This means it’s not an inverted
parabola. And so we can disregard options (C)
and (D). We’ll clear them from the screen to
make more space.
Next, we can find the location of
the 𝑦-intercept by letting 𝑥 equal zero. When we do, our equation becomes 𝑦
equals zero squared minus five times zero plus eight, which is equal to eight. We can see both (A) and (B) have
𝑥-intercepts at eight. So let’s get rid of option (E). Now, our next step might be to find
the location of any 𝑥-intercepts by letting 𝑦 equal zero. But in fact, neither of these
graphs have 𝑥-intercepts. And we’ll investigate that in a
little more detail in a moment. So instead, we’re going to write
our equation in completed square form, in other words, in the form 𝑎 𝑥 plus 𝑘 all
squared plus ℎ. If we can write in that form, then
we know it has a vertex at negative 𝑘, ℎ.
Now the coefficient of 𝑥 squared
is one here, so completing the square is relatively straightforward. We begin by halving the coefficient
of 𝑥, so half of negative five. That’s negative five over two. So in our brackets, we write 𝑥
minus five over two and that’s all squared. We then subtract the square of this
value. So we subtract negative five all
squared. And then we add eight. That’s the same as subtracting 25
over four. And if we write eight as 32 over
four, then we can add these two fractions. Negative 25 over four plus 32 over
four is seven over four. And so in completed square form,
our equation is 𝑥 minus five over two all squared plus seven over four. And so its vertex has coordinates
five over two, seven over four.
Since both the 𝑥- and
𝑦-coordinate here are positive, the vertex of our graph must lie in the first
quadrant. And so the answer is (A) and not
option (B). And at this point, we can
double-check what’s going on with the 𝑥-intercepts. We find these by letting 𝑦 equal
zero. And one way to solve the equation
is to then use completed square form. We subtract seven over four from
both sides, and then we notice that our next step would be to take the square
root. But of course, the square root of a
negative number is not a real value. So there are no real solutions when
we let 𝑦 equal zero, meaning that there are no 𝑥-intercepts. And the answer is therefore option
(A).
What about if we’re given the graph
of a function and asked to find its quadratic equation? This is a very similar process. We’re essentially working
backwards. So in our next example, let’s see
what that would look like.
Write the quadratic equation
represented by the graph shown. Give your answer in factored
form.
Let’s begin by examining the graph
we’ve been given. We might first notice that the
vertex or turning point of this graph has coordinates one, negative nine. This gives us some idea of what the
completed square form equation of this graph might look like. An equation of the form 𝑎 𝑥 plus
𝑘 squared plus ℎ has a vertex negative 𝑘, ℎ. So we let negative 𝑘 be equal to
one and ℎ be equal to negative nine. And we see that the equation of our
graph is 𝑦 equals some constant 𝑎 times 𝑥 minus one all squared minus nine.
So how do we find the value of
𝑎? Well, in fact, we can choose the
coordinate of any point that lies on this curve and substitute that in. A really straightforward one is the
coordinate four, zero. The 𝑥-coordinate is four, and the
𝑦-coordinate is zero. And so our equation becomes zero
equals 𝑎 times four minus one squared minus nine. Well, four minus one squared is
three squared, which is nine. So our equation becomes zero equals
nine 𝑎 minus nine. We add nine to both sides of this
equation. And finally, we’ll divide through
by nine. And when we do, we find that 𝑎 is
equal to one. Substituting this back into the
equation 𝑎 times 𝑥 minus one all squared minus nine, and we find that the equation
of this quadratic is 𝑦 equals 𝑥 minus one squared minus nine.
Now, in fact, we’re told to give
this in factored form. So what next? Well, we’re simply going to
distribute the parentheses, simplify, and then factor. 𝑥 minus one all squared is 𝑥
minus one times 𝑥 minus one. Distributing the parentheses, and
we get 𝑥 squared minus 𝑥 minus 𝑥 plus one. And so, our equation becomes 𝑦
equals 𝑥 squared minus two 𝑥 minus eight. To factor this, we simply find a
pair of numbers that have a product of negative eight and sum to make negative
two. That’s negative four and two. And so, in factored form, the
quadratic equation represented by the graph shown is 𝑦 equals 𝑥 minus four times
𝑥 plus two.
Let’s recap the key concepts from
this lesson. We can sketch the graph of
quadratic equation in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 by first looking at
the coefficient of 𝑥 squared. If it’s positive, the graph is a
parabola. And if it’s negative, it’s an
inverted parabola. In other words, if it’s positive,
then the vertex is its minimum. And if it’s negative, then its
vertex is its maximum. We can set 𝑥 equals zero to give
us the value of the 𝑦-intercept. And we can set 𝑦 equals zero to
find the value of the 𝑥-intercepts if they exist. And if we can write our equation in
completed square form — in other words, 𝑎 times 𝑥 plus 𝑘 all squared plus ℎ — the
vertex has coordinates negative 𝑘, ℎ.
