### Video Transcript

This is the graph of the

exponential function π¦ equals π of π₯. Which of the following is the graph

of π¦ equals four minus π of two π₯?

Letβs compare the original equation

to the equation of our transformed function. Three distinct things have happened

here. Weβve added four, weβve multiplied

the entire function by negative one, and weβve multiplied the π₯ by two. But we need to be careful to ensure

that weβve 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 π¦-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 π of π₯ maps onto π of π₯ plus

π by a translation π units left. Now, if we look at our function,

nothing has actually been added to the value of π₯. And so we arenβt going to be

applying any horizontal translations. But we are going to be applying a

stretch. If we have π of π₯, π of π times

π₯ is a stretch in the horizontal direction by a scale factor of one over π. We have an expression of this

form. In fact, since the coefficient of

π₯ is two, since weβre 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 π¦-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 π¦ equals π of

two π₯ looks as shown, and we can see that thatβs got that horizontal

compression. Itβs 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 π

of two π₯. This is going to result then in a

reflection in the π¦-axis. And so whilst we canβt quite fit

the whole graph on, we know itβs going to pass through negative two on the π¦-axis

as shown. So weβve performed a stretch. Weβve performed a reflection. What about a vertical

translation? This would be of the form π of π₯

plus π. So given an original function π of

π₯, we map it onto π of π₯ plus π by moving it π units up.

If we rewrite our function as π¦

equals negative π of two π₯, which are the transformations weβve 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

π¦-axis at two. And out of our options for (a),

(b), and (c), this is graph (b). So graph (b) is the graph of π¦

equals four minus π of two π₯.