Question Video: Determining Parameters of Functions from their Graphs Mathematics

The graph shows 𝑦 = (π‘˜/(π‘₯ βˆ’ π‘Ž)) + 𝑏. A single point is marked on the graph. What are the values of the constants π‘Ž, 𝑏, and π‘˜?

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Video Transcript

The graph shows 𝑦 equals π‘˜ over π‘₯ minus π‘Ž plus 𝑏. A single point is marked on the graph. What are the values of the constants π‘Ž, 𝑏, and π‘˜?

We begin by noticing that this graph resembles the graph of 𝑦 equals one over π‘₯. We can obtain the given graph from the graph of the parent function 𝑦 equals one over π‘₯ by applying some function transformations. The graph of the parent function 𝑦 equals one over π‘₯ has a horizontal asymptote at 𝑦 equals zero and a vertical asymptote at π‘₯ equals zero. The given graph has a horizontal asymptote at 𝑦 equals negative two and a vertical asymptote at π‘₯ equals three. This means that a downward shift of two units and a rightward shift of three units is one of the function transformations used to obtain this graph from the graph of 𝑦 equals one over π‘₯.

Before applying this translation, however, we first need to check if any other transformations are involved, since the order of transformations is very important. There are three different types of transformations to consider: translation, dilation, and reflection. The given graph is oriented the same way as the parent function, so we can rule out reflection. Dilation is a possibility, but it’s difficult to judge by eye. Since we are given the point six, negative one on the graph, we can use this point to determine the dilation factor, if there is one. Recall that when combining transformations, dilations and reflections must be done before translations.

Recall that a horizontal dilation by a scale factor of 𝑑 one means that we map π‘₯ to π‘₯ over 𝑑 one. In other words, the π‘₯-values of all the points on the graph are reduced by a factor of 𝑑 one. A vertical dilation by a scale factor of 𝑑 two means mapping 𝑓 of π‘₯ to 𝑑 two times 𝑓 of π‘₯. In other words, we multiply all of the 𝑦-values on the graph by a factor of 𝑑 two. Beginning with the parent function 𝑓 of π‘₯ equals one over π‘₯ and performing a horizontal dilation, we get one over π‘₯ over 𝑑 one.

Next, performing a vertical dilation, we multiply this new 𝑓 of π‘₯ by 𝑑 two, giving 𝑑 two over π‘₯ over 𝑑 one. This simplifies to 𝑑 one 𝑑 two over π‘₯. 𝑑 one times 𝑑 two is just another constant in and of itself. So, we can call this constant π‘˜. We do not yet know the value of π‘˜. First, we need to move on to translations. The positions of asymptotes are unaffected by dilations. Therefore, these could only have been moved by the translation of two units downwards and three units to the right. A vertical translation of 𝑐 units means mapping 𝑓 of π‘₯ to 𝑓 of π‘₯ plus 𝑐. So, we change all of the 𝑦-values of the points on the graph from 𝑦 to 𝑦 plus 𝑐. 𝑐 being positive means the graph moves upwards and negative means it moves downwards.

In our case, we need to perform the dilation first. So, the parent function one over π‘₯ goes to π‘˜ over π‘₯. Then, to translate this by two units downwards, this goes to π‘˜ over π‘₯ minus two. Now, recall that for a horizontal translation by 𝑐 units, we map all of the π‘₯-values to π‘₯ minus 𝑐. Again, the value of 𝑐 will determine the direction, with positive 𝑐 giving a rightward shift and negative 𝑐 giving a leftward shift. But of course, the sign will be reversed since we are subtracting 𝑐 from π‘₯. So, starting from π‘˜ over π‘₯ minus two, to perform a rightward shift of three units, we need to change the π‘₯-values to π‘₯ minus three. So, this gives us π‘˜ over π‘₯ minus three minus two. So, this graph represents the function 𝑓 of π‘₯ equals π‘˜ over π‘₯ minus three minus two. So, we have found two of the parameters in the question. π‘Ž is equal to three and 𝑏 is equal to negative two.

Now we just need to find the value of π‘˜. This equation has three unknowns in it. We have 𝑓 of π‘₯, π‘₯, and π‘˜. We are given a point on the graph: six, negative one. And this corresponds to values of 𝑓 of π‘₯ and π‘₯, respectively, which we can then substitute into the equation and rearrange to find π‘˜. So, we have 𝑓 of π‘₯ equals negative one and π‘₯ equals six. Simplifying and rearranging gives π‘˜ equals three. So, we now have the values of all three constants. π‘Ž is equal to three, 𝑏 is equal to negative two, and π‘˜ is equal to three.

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