### Video Transcript

Which of the following is the graph
of π of π₯ is equal to π₯ plus one all squared times π₯ minus two multiplied by π₯
minus one? Option (A), (B), (C), (D), or
(E).

Weβre given five possible sketches
of the curve π¦ is equal to π of π₯. And we could see if we were to
multiply our expression for π of π₯, we would get a polynomial. So to answer this question, weβll
use what we know about sketching polynomials. The first thing we can do when
weβre sketching any function is substitute π₯ is equal to zero to find the
π¦-intercept. So weβll substitute π₯ is equal to
zero into our function π of π₯. So we substitute π₯ is equal to
zero into π of π₯. We get zero plus one all squared
times zero minus two multiplied by zero minus one. And if we evaluate this expression,
we get two.

Now, two is a positive number. So our π¦-intercept must be above
the π₯-axis. And if we look in option (E), we
can see that the π¦-intercept is below the π₯-axis. It canβt be option (E). The second thing we should look for
is the end behavior of our curve. For polynomial functions, we can
determine the end behavior of a curve by looking at the leading term. In our case, weβre not given an
expanded form for our graph. So we donβt know the leading
term. We would need to multiply this
out.

However, we only need the highest
power of π₯. If we were to do this, we would see
we would just get π₯ to the fourth power. So we can determine the end
behavior of our curve π¦ is equal to π of π₯ just by looking at the end behavior of
π₯ to the fourth power. We know for large positive values
of π₯, π₯ to the fourth power is positive. And for very negative values of π₯,
itβs also positive.

Also, remember, we showed that the
π¦-intercept must be at two. So we can clear this to avoid
confusion and add this to our sketch. But now we can see something
interesting. Options (B), (C), and (D) donβt
have the same end behavior as our curve. So none of these can be the correct
sketch. Therefore, by elimination, option
(A) must be the correct sketch. Itβs also worth pointing out we
couldβve sketched this directly from the function π of π₯.

The third thing we want to do when
sketching a polynomial curve is find the π₯-intercepts. π₯-intercepts will be where our
functionβs output is equal to zero. And if we factor our polynomial,
the products of factors being equal to zero means that one of our factors must be
equal to zero. So we just solve each factor equal
to zero. And this gives us our three
π₯-intercepts: negative one, two, and one. And we can add these to our
sketch.

The last thing we need to know is
the shape of our function near the intercepts. And we recall we can do this by
looking at the multiplicity of the factor. For example, when π₯ is equal to
two and when π₯ is equal to one, the multiplicity of these two factors is one. We could find a direct
approximation by substitution. However, this is not necessary. We know this is going to behave
like a linear function, a straight line.

Look at π₯ is equal to two
however. We know the end behavior of our
curve near this point. So this slope at π₯ is equal to two
must be positive. It must be a positive linear
function. But then the opposite must be true
at one, because our curve canβt intersect the π₯-axis anywhere else. It must be a negative slope linear
function. And this means we can fill in some
of our curve. It will look something like
this.

When π₯ is equal to negative one
however, when we look at our factor, we can see its multiplicity is two. This means near the value of
negative one, our curve will look like some multiple of π₯ plus one all squared. Now, we could find the value of π΄
by substituting negative one into the remaining part of our function. However, itβs not necessary. We know that π΄ must be positive
just by looking at our curve. This is the only way that the shape
of our curve could make sense. So π΄ is positive. And we can fill in the rest of our
curve. Our curve is just going to touch
the π₯-axis at the point negative one. And this confirms that option (A)
was the correct choice.