Video Transcript

This is the graph of the
exponential function 𝑦 equals 𝑓 of π‘₯. Which of the following is the graph
of 𝑦 equals four minus 𝑓 of two π‘₯?

Let’s compare the original equation
to the equation of our transformed function. Three distinct things have happened
here. We’ve added four, we’ve multiplied
the entire function by negative one, and we’ve multiplied the π‘₯ by two. But we need to be careful to ensure
that we’ve done these in the correct order. The order is as follows. We begin by looking for any
horizontal translations. Is the graph going to move left or
right? Then we stretch it, and we can do
that in both directions if necessary. We then perform any reflections
either across the π‘₯- or 𝑦-axis or both. And finally, we deal with any
vertical translations. Does the graph move up or down?

So a horizontal translation occurs
when we add some constant to the value of π‘₯. So 𝑓 of π‘₯ maps onto 𝑓 of π‘₯ plus
π‘Ž by a translation π‘Ž units left. Now, if we look at our function,
nothing has actually been added to the value of π‘₯. And so we aren’t going to be
applying any horizontal translations. But we are going to be applying a
stretch. If we have 𝑓 of π‘₯, 𝑓 of 𝑏 times
π‘₯ is a stretch in the horizontal direction by a scale factor of one over 𝑏. We have an expression of this
form. In fact, since the coefficient of
π‘₯ is two, since we’re multiplying our value of π‘₯ by two, we have a horizontal
stretch by a scale factor of one-half. We can add this interim graph to
our diagram.

We know it will still pass through
the 𝑦-axis at two. Then the π‘₯-value of the point with
coordinates one, approximately 0.7 will half, and we get the coordinate 0.5,
0.7. Similarly, the curve passes through
the point two and approximately 0.2. The π‘₯-value will halve at this
coordinate. And we get one, 0.2. Up here, we have a point which has
coordinates negative 0.6 roughly and four. And halving the π‘₯-coordinate, that
becomes negative 0.3, four. And so the graph of 𝑦 equals 𝑓 of
two π‘₯ looks as shown, and we can see that that’s got that horizontal
compression. It’s a stretch by a scale factor of
one-half. And what about reflections? Well, these occur when we either
change the coefficient of π‘₯ to be negative or we multiply the whole function by
negative one.

Well, we do indeed have negative 𝑓
of two π‘₯. This is going to result then in a
reflection in the 𝑦-axis. And so whilst we can’t quite fit
the whole graph on, we know it’s going to pass through negative two on the 𝑦-axis
as shown. So we’ve performed a stretch. We’ve performed a reflection. What about a vertical
translation? This would be of the form 𝑓 of π‘₯
plus 𝑑. So given an original function 𝑓 of
π‘₯, we map it onto 𝑓 of π‘₯ plus 𝑑 by moving it 𝑑 units up.

If we rewrite our function as 𝑦
equals negative 𝑓 of two π‘₯, which are the transformations we’ve performed, plus
four, we should now be able to see that we need to move our graph four units up. And so it will now pass through the
𝑦-axis at two. And out of our options for (a),
(b), and (c), this is graph (b). So graph (b) is the graph of 𝑦
equals four minus 𝑓 of two π‘₯.

You are watching: Question Video: Combined Translations and Stretches of Graphs. Info created by GBee English Center selection and synthesis along with other related topics.