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Video: Combined Transformations of Graphs

Bethani Gasparine

The following graph is a transformation of the graph of y = x . What is the function it represents? Write your answer in a form related to the transformation. <Figure>

02:29

Video Transcript

The following graph is a transformation of 𝑦 equals the absolute value of π‘₯. What is the function it represents? Write your answer in a form related to the function.

Here, we’re looking at the general 𝑦 equals the absolute value of π‘₯ with some transformations. π‘Ž, first looking at π‘Ž and what it would do to the absolute value graph. If it would be positive, the graph would look like a V; it would be facing upwards. If π‘Ž were negative, it would be the exact opposite; it would be opening downwards. If the absolute value of π‘Ž, meaning we’re not looking at the sign anymore, if the absolute value of π‘Ž is greater than one, this will sh- vertically stretch the graph. If π‘Ž is between zero and one, so if it’s a fraction, it will be a vertical compression of the graph.

Now looking at β„Ž, that will be the horizontal shift, so moving left and right. And finally, π‘˜ will be the vertical shift, how much it moved up or down.

So first, looking at our graph, it is upside down. So that means π‘Ž must be negative. And next, to determine what π‘Ž actually is, so if it’s β€” the absolute value will be bigger than one or if it will be between zero and one, it’s essentially like the slope of the line. And we can see, if we go from the peak to the right, it goes down one, right one. So π‘Ž is one, which this kind of makes sense. Looking at that, it’s negative one; it’s the slope.

Next, we look at β„Ž, the horizontal shift. So it’s π‘₯ minus the horizontal shift. So the original 𝑦 equals the absolute value of π‘₯ graph would start at zero. So-so if the original graph started at zero, zero, our new graph started one space to the left, which would be a negative one shift.

And then finally, how much would it have moved up? It went up four.

So to simplify this, we have 𝑦 equals negative one absolute value π‘₯ plus one plus four. We could use the commutative property of addition because we’re adding four to that absolute value. So we could put the four first and then put minus. We don’t have to put the one in front of the absolute value. So we could have 𝑦 equals four minus the absolute value of π‘₯ plus one.