Video Transcript
Which of the following graphs
represents the equation 𝑦 equals two 𝑥 squared minus three 𝑥 plus one?
We’re given some graphs of
quadratic functions, and we want to determine which of these represents the equation
𝑦 equals two 𝑥 squared minus three 𝑥 plus one.
To do this, we recall that a
quadratic function has the form 𝑓 of 𝑥 is equal to 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐
for some constants 𝑎, 𝑏, and 𝑐, where 𝑎 is nonzero. And we can note various properties
about quadratics from this general form. The first thing is that all
quadratics have a parabolic shape. And if the constant 𝑎, which we
call the leading coefficient, is less than zero, our quadratic opens downward,
whereas if 𝑎 is greater than zero, the graph will open upward.
In our equation 𝑦 equals two 𝑥
squared minus three 𝑥 plus one, we see that 𝑎 is equal to two. That’s the coefficient of 𝑥
squared. And this is greater than zero. So the graph representing our
function must open upwards. This means we can eliminate graphs
(D) and (E) since these both open downwards.
We also know that all graphs of
quadratics have a 𝑦-intercept at the point zero, 𝑐. This is the point where the graph
crosses the 𝑦-axis. Now, in our case, we see that 𝑐 is
equal to one, so that our 𝑦-intercept must be the point with coordinates zero,
one. And if we consider our three
remaining options (A), (B), and (C), we see that graph (A) has its 𝑦-intercept at
the point zero, negative one. So we can eliminate graph (A) since
this does not match the value of 𝑐 in our equation.
Both of graphs (B) and (C) have
their 𝑦-intercept at the point zero, one. So these are both still contenders
for the given equation. Something else we know about
quadratic functions is that they can have either zero, one, or two
𝑥-intercepts. And the 𝑥-coordinates of these
points tell us the input values for which the function outputs zero.
Now we see that in both of our
remaining graphs, we have two 𝑥-intercepts. So let’s look at the coordinates of
one of these 𝑥-intercepts for each of our two graphs to see if they satisfy the
given equation. Taking the 𝑥-intercept with
coordinates one, zero from graph (B), by substituting 𝑥 equals one and 𝑦 equals
zero into our equation, we can see if the right-hand and left-hand sides of our
equation match for this point. On our right-hand side, we have two
times one squared minus three times one plus one, which is two minus three plus
one. And this is equal to zero. So our point one, zero does satisfy
the given equation.
Now, trying the same thing for one
of the 𝑥-intercepts in graph (C), taking the point negative one, zero and
substituting these coordinates into our equation, on our right-hand side, we have
two times negative one squared minus three times negative one plus one. Since negative one squared is equal
to one and negative three times negative one is equal to three, this gives two plus
three plus one, which is equal to six.
Our right- and left-hand sides are
not equal for this point. So this point does not satisfy the
given equation. And since this point lies on graph
(C), graph (C) cannot represent the given equation.
This leaves us with graph (B). We know that graph (B) is the
correct form for the given equation, that is, it opens upward, that its 𝑦-intercept
satisfies the equation, and that one of the 𝑥-intercepts also satisfies the
equation. For added confirmation that graph
(B) does represent the given equation, we could check that some other points on the
graph satisfy the equation. So let’s do this.
It’s sensible to check our second
𝑥-intercept, which has coordinates one-half, zero. So, if we substitute these values
into our equation, on the right-hand side, we have two times one-half squared minus
three times one-half plus one. And we want to know if this is
equal to zero. In fact, this does evaluate to
zero. Hence, this second 𝑥-intercept of
graph (B) does satisfy the given equation.
As a final confirmation, we see
that the point with coordinates two, three is also on the graph in option (B). So let’s try this in the given
equation. Substituting 𝑥 is equal to two and
𝑦 equals three into the equation gives on our right-hand side two times two squared
minus three times two plus one, and three on our left-hand side. The right-hand side evaluates to
eight minus six plus one, which is equal to three. So our right- and left-hand sides
are equal, and the point does satisfy the given equation.
We find then that since the leading
coefficient of the function is positive and the 𝑥- and 𝑦-intercepts and a further
point on the graph satisfy the given equation, graph (B) represents the equation 𝑦
equals two 𝑥 squared minus three 𝑥 plus one.
