### Video Transcript

Sketch the graph of π of π₯ equals
three π₯ squared over π₯ squared minus four.

Letβs begin by considering the
domain and range of our function. We know that a function which is a
quotient wonβt exist at points where the denominator of the quotient is zero. So we set π₯ squared minus four
equals zero and solve for π₯ to find the domain of our function. We can add four to both sides of
this equation. Then we take the square root of
both sides of our equation, remembering to take both the positive and negative
square root of four. And we obtain that when π₯ squared
minus four is equal to zero, π₯ is equal to positive or negative two. So the domain of our function is
all real values, except π₯ is equal to positive or negative two. Now the range itself is a set of
possible outputs on the function. And we can infer this through the
graphing process.

Next, weβll work out whether there
are any intercepts. By setting π₯ equal to zero and
solving for π¦, weβll find the π¦-intercept. When we do, we obtain π¦ to be
equal to three times zero squared over zero squared minus four, which is zero. So thereβs a π¦-intercept at π¦
equals zero. Next, we set π¦ or π of π₯ equal
to zero and solve for π₯. That is, zero equals three π₯
squared over π₯ squared minus four.

Now for this to be true, we know
that the numerator of this fraction must itself be equal to zero. And for three π₯ squared to be
equal to zero, π₯ must be equal to zero. So we actually only have one
intercept altogether. And thatβs at the origin: zero,
zero. Next, we can check for
symmetry. An even function is one for which
π of negative π₯ equals π of π₯, whereas an odd function is one for which π of
negative π₯ equals negative π of π₯. Well, π of negative π₯ is three
times negative π₯ squared over negative π₯ squared minus four, which is equal to π
of π₯. So our function is even. And that means itβs going to be
symmetrical about the π¦-axis.

Next, weβll look for
asymptotes. Weβll look for horizontal
asymptotes by considering what happens to our function as π₯ approaches either
positive or negative infinity. We canβt use direct substitution as
when we substitute in π₯ equals either positive or negative infinity, we end up with
infinity over infinity, which is undefined. So instead, we divide both the
numerator and the denominator of our expression by π₯ squared. That gives us three over one minus
four over π₯ squared.

And now we can use direct
substitution. As π₯ approaches either positive or
negative infinity, four over π₯ squared approaches zero. So the limit of our function is
three over one, which is just three. And we see that the line π¦ equals
three must be a horizontal asymptote.

Remember, we said that when π₯ is
equal to positive or negative two, the denominator is zero. So we find the following
limits. The limit as π₯ approaches two from
the right of our function is infinity. And the limit as π₯ approaches two
from the left of the function is negative infinity. And the limit as π₯ approaches
negative two from the right of the function is negative infinity. And as it approaches negative two
from the left, itβs positive infinity. So we obtain π₯ equals positive or
negative two to be vertical asymptotes.

Letβs clear some space and look for
intervals of increase and decrease. Weβll begin by using the quotient
rule to find the first derivative of our function. By letting π’ be equal to three π₯
squared and π£ be equal to π₯ squared minus four, we obtain expressions for dπ’ by
dπ₯ and dπ£ by dπ₯. And so the first derivative of our
function is π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ over π£ squared. And distributing our parentheses,
we find the first derivative to be negative 24π₯ over π₯ squared minus four all
squared.

Now since π₯ squared minus four all
squared must be greater than zero for all π₯, this means the first derivative of our
function must be less than zero when π₯ is greater than zero. So for values of π₯ greater than
zero, we get negative 24 multiplied by a positive, which gives us a negative
value. And so for π₯ greater than zero,
the first derivative is less than zero. The opposite is true. When π₯ is less than zero, the
first derivative is greater than zero. Since we know the function doesnβt
exist when π₯ is equal to negative two, we can say that its intervals of increase
are negative infinity to negative two and negative two to zero. And it has open intervals of
decrease over zero to two and two to infinity.

Now, actually, if we look
carefully, we also see that the first derivative is equal to zero when π₯ is equal
to zero. So thereβs a critical point at the
point zero, zero. Coincidently, thatβs also the point
where our curves cross the axes.

We now have everything we need to
complete our sketch. We know that thereβs a horizontal
asymptote at π¦ equals three and vertical asymptotes at π₯ equals negative two and
π₯ equals two. Thereβs a critical point and an
intercept at zero, zero. And thatβs important because we see
that the curve wonβt be able to cross the π₯-axis here or here. That helps us decide where these
two parts of the curve go. The curve is increasing over the
open interval negative infinity to negative two and decreasing over the open
interval two to infinity. And of course, we have these
asymptotes. The function is also even and
symmetrical about the π¦-axis. So this confirms weβre probably on
the right track. We then use the remaining pieces of
information to sketch the final part of our curve. And weβve successfully sketched the
graph of π of π₯ equals three π₯ squared over π₯ squared minus four.