Question Video: Identifying the Graph of a Transformation of the Sine Function Mathematics

Identify the graph of the function 𝑓(π‘₯) = 1/2 sin π‘₯.

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Video Transcript

Identify the graph of the function 𝑓 of π‘₯ equals a half sin π‘₯.

The function whose graph we’ve been asked to identify is a transformation of the graph of the sine function. We’ll begin by recalling what the graph of the sine function itself looks like, and then we’ll consider how it has been transformed.

We should recall the general shape of the sine function: a smooth curve oscillating between its minimum and maximum values. Those minimum and maximum values are negative one and positive one, respectively. The function is periodic with a period of 360 degrees or two πœ‹ radians. So the same pattern repeats after every interval of this length. The graph of the sine function passes through the origin. So sin of zero degrees or sin of zero radians is zero. The roots of the sine function occur at every integer multiple of πœ‹. So that’s πœ‹, two πœ‹, three πœ‹, four πœ‹ radians, and so on.

We’ve now identified what the graph of the sine function looks like. The function whose graph we need to identify though is 𝑓 of π‘₯ equals a half sin π‘₯. So the sine function has been multiplied by one-half. We recall that multiplying an entire function by a constant corresponds to a vertical dilation or stretch of the function by that scale factor. So here the sine function has been stretched vertically by a scale factor of one-half. This means that the maximum value of 𝑓 of π‘₯ will be positive one-half instead of one and the minimum value will be negative one-half. There’s been no change to the variable, so there’s no horizontal transformation and therefore no change to the periodicity of the graph. We can therefore sketch the graph of 𝑓 of π‘₯ on the same axes as our sine function.

We now know that we’re looking for a graph with a maximum value of one-half and a minimum value of negative one-half. Considering options (A) and (B) and looking carefully at the vertical scale, we can rule out each of these graphs because they each oscillate between a minimum value of negative two and a maximum value of positive two. We can also rule out graph (E) because this graph continues indefinitely in the vertical direction. In fact, graph (E) is the graph of the tangent function.

We’re left then with graphs (C) and (D), both of which have a minimum value of negative a half or negative 0.5 and a maximum value of 0.5. They also each have the correct period of two πœ‹ radians. To distinguish between the two graphs, we need to consider the roots. Looking at graph (C), we can see that this does pass through the origin, and it has its roots at every integer multiple of πœ‹ radians. Graph (D), on the other hand, doesn’t pass through the origin, and its roots are instead at πœ‹ by two, three πœ‹ by two, negative πœ‹ by two, negative three πœ‹ by two, and so on. We can therefore rule out graph (D) as it has the incorrect roots.

The graph of the function 𝑓 of π‘₯ equals a half sin π‘₯, which is a vertical stretch of the graph of sin π‘₯ with a scale factor of one-half, is graph (C).

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