Video Transcript
Find the function shown in the
figure. Is it (A) π¦ equals negative π₯
minus four cubed plus four? (B) π¦ equals π₯ minus four cubed
plus four. (C) π¦ equals negative π₯ plus four
cubed minus four. Or is it (D) π¦ equals π₯ plus four
cubed minus four?
The first thing to note when
answering this question is that the graph is in the shape of a cubic function. This can be confirmed by noticing
that all of the options given have powers of three on the term with π₯, so they are
all of order three. For comparison, let us draw the
graph of π₯ cubed, which is the simplest cubic graph possible. To do this, we plot some of the
points that the curve must pass through and join them up with a curve. So we plot negative two, negative
eight; negative one, negative one; zero, zero; one, one; and two, eight. We should also make sure that the
curve is flat at the origin due to this shape of the cubic function.
We note that compared to the given
function, the π₯ cubed function is orientated in the opposite direction. Additionally, the curves have been
translated downwards and to the left. This can be observed by noticing
that the point that was at the origin is now at the points negative four, negative
four. If we were to summarize what has
happened to the graph in terms of transformations, we can break it down into three
steps. Let us list these out. First of all, the graph has been
reflected in the π¦-axis, or equivalently in the π₯-axis. Next, this is followed by a
translation of four units downwards. Finally, the curve is shifted four
units to the left.
We can also verify that these
translations are correct by considering the movement of the point of inflection from
zero, zero to negative four, negative four. That is to say, if we move zero,
zero downwards by four units and left by four units, we do indeed arrive at the new
inflection point.
Now, if we start from the equation
π¦ equals π₯ cubed, we must ask the question of what these transformations do to the
equation. First of all, the reflection in the
π¦-axis leads to the sign in front of the function changing. Thus, the equation becomes π¦
equals negative π₯ cubed. Alternatively, we note that if we
treated this as a reflection in the π₯-axis instead, we would swap the sign of π₯ to
negative. However, since π₯ cubed is an odd
function, this would result in the same effect on the equation.
Next, we consider the effect of
shifting the graph down four units. This is achieved by subtracting
four from the equation. Therefore, the new equation will be
π¦ equals negative π₯ cubed minus four.
Lastly, a translation to the left
by four is equivalent to adding four to the input value π₯. When applying this to our equation,
we should be careful to make sure to apply it to the π₯-variable itself and not π
of π₯. If we do this, we get π¦ equals
negative π₯ plus four cubed minus four.
Finally, comparing this equation to
our options, we see that this corresponds to option (C).
