### Video Transcript

Sketch the graph of the
vector-valued function π of π‘ equals π‘ cubed π plus π‘ π.

Letβs begin by recalling what this
vector-valued function actually tells us. It takes a real number π‘, and it
outputs a position vector. Itβs horizontal component is π‘
cubed and its vertical component is π‘. So we could say that in the
π₯π¦-plane, the π₯-value of any coordinate on our graph would be given by π‘ cubed
and the π¦-value would be given by π‘. And there are two things we could
do next. We could try forming a table and
inputting values of π‘ and plotting the π₯- and π¦-coordinates. Alternatively, we can manipulate
our equations for π₯ and π¦ to see if we can eliminate our parameter and get
something we recognise. Letβs look at that latter
method.

We are told that π¦ is equal to
π‘. We can, therefore, replace π‘ with
π¦ in our equation for π₯. And we find that π₯ is equal to π¦
cubed. We could alternatively say that π¦
is equal to the cube root of π₯. So how do we sketch this graph? Well, we know how to sketch the
graph of π¦ equals π₯ cubed. But we also know that π¦ is equal
to the cube root of π₯ is the inverse function of π¦ equals π₯ cubed. We then recall that to sketch the
graph of an inverse function, we reflect the graph of the original function in the
line π¦ equals π₯. And we obtain the graph of π¦ is
equal to the cube root of π₯ or π₯ is equal to π¦ cubed, as shown.

Now, of course, we arenβt actually
quite finished. Remember, we created a pair of
parametric equations. And we know that when we plot a
parametric graph, we must consider the direction in which the curve is sketched. So letβs take a couple of
values. Letβs consider π‘ equals zero and
π‘ equals one. When π‘ is equal to zero, π₯ is
equal to zero cubed or zero and π¦ is equal to zero. Similarly, when π‘ is equal to one,
π₯ is equal to one cubed, which is, of course, one and π¦ is also equal to one. So we start at the point zero, zero
and we move up to the point one, one. This means weβre moving on this
curve from left to right. And now, weβve finished. Weβve sketched the graph of the
vector-valued function π of π‘ equals π‘ cubed π plus π‘ π.