Question Video: Describing the End Behavior of a Function from Its Graph Mathematics

Consider the graph of the function 𝑓(π‘₯) = 1/(1 βˆ’ π‘₯) + 2. What is the end behavior of the graph as π‘₯ approaches 1? [A] The value of 𝑦 approaches ∞ when π‘₯ gets closer to 1 from the positive direction and approaches βˆ’βˆž when π‘₯ gets closer to 1 from the negative direction. [B] The value of 𝑦 approaches βˆ’βˆž when π‘₯ gets closer to 1 from the positive direction and approaches ∞ when π‘₯ gets closer to 1 from the negative direction. [C] The value of 𝑦 approaches ∞ when π‘₯ gets closer to 1 from the negative direction or from the positive direction. [D] The value of 𝑦 approaches βˆ’βˆž when π‘₯ gets closer to 1 from the negative direction or the positive direction.

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Video Transcript

Consider the graph of the function 𝑓 of π‘₯ is equal to one divided by one minus π‘₯ plus two. What is the end behavior of the graph as π‘₯ approaches one? Option (A) the value of 𝑦 approaches ∞ when π‘₯ gets closer to one from the positive direction and approaches negative ∞ when π‘₯ gets closer to one from the negative direction. Option (B) the value of 𝑦 approaches negative ∞ when π‘₯ gets closer to one from the positive direction and approaches ∞ when π‘₯ gets closer to one from the negative direction. (C) The value of 𝑦 approaches ∞ when π‘₯ gets closer to one from the negative direction or from the positive direction. Or is it option (D) the value of 𝑦 approaches negative ∞ when π‘₯ gets closer to one from the negative direction or the positive direction?

In this question, we’re given the graph of a function: 𝑓 of π‘₯ is equal to one divided by one minus π‘₯ plus two. And we need to use this graph or the given function to determine the end behavior of this graph as the values of π‘₯ approach one. And there’s many different ways we could go about answering this question. For example, we can determine the end behavior of this graph as π‘₯ approaches one entirely by using the given function.

To see how we might do this, let’s consider the denominator in the given function: one minus π‘₯. If our values of π‘₯ are greater than one, which means π‘₯ is in the positive direction of one, then we can note one minus π‘₯ will be negative, since we’re subtracting a number bigger than one from one. We can also note as the values of π‘₯ get closer and closer to one, the magnitude of one minus π‘₯ will get closer and closer to zero. In other words, the distance between one and π‘₯ gets smaller as π‘₯ gets closer and closer to one. So we’re dividing one by a negative number with a very small magnitude. And as the magnitude of this number gets smaller, its reciprocal will get larger in magnitude. And adding two to this value won’t change this fact. So 𝑓 of π‘₯ will approach negative ∞ as our values of π‘₯ approach one from positive direction.

However, this level of analysis is very difficult. So, instead, it’s very useful to be able to do this from a given diagram or, if possible, to sketch a diagram to help us determine this information. We can do this by recalling the input values of the function are the π‘₯-coordinates of the point on the curve and the output values are the corresponding 𝑦-coordinates. This means we can determine what happens to the outputs of the function as our values of π‘₯ approach one from the positive direction by seeing what happens to the 𝑦-coordinates of points which lie on the curve.

To do this, let’s add in the vertical line π‘₯ is equal to one onto our diagram. We can see that the curve gets closer and closer to this line. However, it never touches the line. This is a vertical asymptote of the curve. In particular, we can see as the values of π‘₯ approach one from the right, the 𝑦-coordinates of the points on the curve are going lower and lower. They’re unbounded, so they’re approaching negative ∞. And we get a similar story if we look at the values of π‘₯ approaching one from the negative direction. We can see that the 𝑦-coordinates of the point on the curve are getting larger and larger, and they’re unbounded. So they’re approaching positive ∞.

And we can see that this only matches option (B). The value of 𝑦 approaches negative ∞ when π‘₯ gets closer to one from the positive direction and approaches ∞ when π‘₯ gets closer to one from the negative direction.

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