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In this video, we will learn how to graph rational functions whose denominators are linear, determine the types of their asymptotes, and describe their end behaviors.
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A rational function is an algebraic fraction where both the numerator and denominator are polynomials.
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For example, π of π₯ equal to five π₯ plus seven over two π₯ minus one is a rational function.
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Itβs important to note that a regular polynomial function, for example, π of π₯ equals three π₯ minus five, is also a rational function.
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We could rewrite three π₯ minus five as three π₯ minus five over one, and one is a degree zero polynomial.
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However, regular polynomial functions like π of π₯ are slightly different to other rational functions.
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This is because any value of π₯ is valid in the domain of π, whereas a rational function with a degree one polynomial on its denominator, like this one, will have one value of π₯ for which the polynomial is equal to zero.
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And therefore, the function π of π₯ is undefined.
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In this case, if two π₯ minus one is equal to zero, then π₯ is equal to one-half.
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Therefore, the value of one-half must be excluded from the domain of π because it is undefined at this value.
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The simplest rational function with a non-zero-degree polynomial on its denominator is the reciprocal function π of π₯ equals one over π₯.
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The graph of the function π¦ equals one over π₯ forms the shape of a hyperbola, which looks something like this.
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The hyperbola is symmetric about the lines π¦ equals π₯ and π¦ equals negative π₯.
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In the first quadrant, the curve approaches the line π¦ equals zero as π₯ tends to β without ever actually touching it.
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The line π¦ equals zero or the π₯-axis is called an asymptote.
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The curve also approaches the line π¦ equals zero in the third quadrant, this time from below as π₯ tends to negative β.
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Similarly, the curve approaches β as π₯ tends to zero from the positive side and negative β as π₯ tends to zero from the negative side.
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The graph of π¦ equals one over π₯ therefore has a horizontal asymptote at π¦ equals zero and a vertical asymptote at π₯ equals zero.
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We can use this most basic function π¦ equals one over π₯ and apply function transformations to obtain different graphs of different rational functions.
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In the first example, we will identify the graph of a rational function using a function transformation from the graph of π¦ equals one over π₯.
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Which of the following graphs represents π of π₯ equals one over π₯ plus one?
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Letβs start by recalling the graph of the simplest rational function π¦ equals one over π₯.
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This graph has a horizontal asymptote at π¦ equals zero and a vertical asymptote at π₯ equals zero.
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We can obtain the graph of π of π₯ equals one over π₯ plus one by using a function transformation on π¦ equals one over π₯.
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To transform the function one over π₯ into one over π₯ plus one, we simply let π₯ map to π₯ plus one.
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Now, recall that a function transformation of π₯ to π₯ plus π corresponds to a horizontal shift to the left by π units.
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Therefore, the graph of π¦ equals one over π₯ plus one is the same as the graph of π¦ equals one over π₯ shifted to the left by one unit.
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Looking at the possible answers, we can see that graph (c) is the same as the graph of π¦ equals one over π₯ shifted to the left by one unit.
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Its vertical asymptote π₯ equals zero has shifted to the left to π₯ equals negative one.
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And its horizontal asymptote at π¦ equals zero is unchanged.
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Now, letβs look at the opposite kind of example where we are given a graph and asked to find the function.
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What function is represented in the figure below?
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To begin with, notice that the given graph resembles the graph of π¦ equals one over π₯.
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We can obtain this graph from the graph of the parent function π¦ equals one over π₯ by applying a few function transformations.
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The graph of π¦ equals one over π₯ has a horizontal asymptote at π¦ equals zero and a vertical asymptote at π₯ equals zero.
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The given graph also has a vertical asymptote at π₯ equals zero, but its horizontal asymptote is at π¦ equals negative three.
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This means that a downward shift of three units is one of the function transformations used to obtain the given graph from the graph of the parent function.
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Before we apply this vertical shift, we first need to check if any other function transformations are involved, since the order of function transformations is very important.
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There are three different types of transformations to consider: translation, dilation, and reflection.
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We have already seen that there is a translation of three units downwards.
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Looking at these two points indicated on the given graph, if we translate these two points back up three units, then we get the points negative one, one and one, negative one.
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These two points clearly are not on the graph of the parent function π¦ equals one over π₯.
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Therefore, there must have been at least one other transformation involved.
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They do look similar to the points negative one, negative one and one, one.
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The two points are the same distance apart and have the same absolute values for their π₯- and π¦-coordinates.
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Therefore, there doesnβt seem to have been any dilation involved, which would either increase or decrease the distance between the points.
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The given graph appears to be upside down when compared with the parent function.
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More specifically, the value of π¦ approaches positive β as π₯ tends to zero from the negative direction.
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And it approaches negative β as π₯ tends to zero from the positive direction.
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This is exactly the opposite behavior of the graph of the parent function, which implies that there is a reflection involved.
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Due to the symmetry of the graph of the parent function π¦ equals one over π₯, a reflection over the π₯-axis is the same as a reflection over the π¦-axis.
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So, for default, we will say this is a reflection over the π₯-axis.
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When combining transformations, dilations and reflections must be done before translations.
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A reflection over the π₯-axis means we swapped the sign of the π¦-value of all of the points on the graph.
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So, π of π₯ goes to negative π of π₯.
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This implies that we multiply the original function by negative one.
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So, so far, we are taking the parent function one over π₯ and multiplying it by negative one to give negative one over π₯.
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Letβs clear a little space, and letβs look at the graph of what we have so far: π¦ equals negative one over π₯.
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The graph has been reflected in the π₯-axis.
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And the point negative one, negative one has become negative one, one, and the point one, one has become one, negative one.
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These are the two points that we obtained on the given graph when shifting it up by three units.
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So, now, all we need to do is apply the translation of three units downwards.
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A vertical translation of π units means adding π to the π¦-values of all of the points on the graph.
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So, π of π₯ goes to π of π₯ plus π.
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And the sign of π will indicate the direction the graph moves in.
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If π is positive, it will move upwards, and if negative, it will move downwards.
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Therefore, to translate this graph three units downwards, we subtract three from π of π₯.
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So, negative one over π₯ goes to negative one over π₯ minus three.
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Therefore, the given graph represents the function π of π₯ equals negative one over π₯ minus three.
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And we can substitute in the original two points negative one, negative two and one, negative four to show that they do indeed satisfy this equation.
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In the next example, we will determine missing parameters in a rational function from the given graph.
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The graph shows π¦ equals π over π₯ minus π plus π.
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A single point is marked on the graph.
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What are the values of the constants π, π, and π?
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We begin by noticing that this graph resembles the graph of π¦ equals one over π₯.
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We can obtain the given graph from the graph of the parent function π¦ equals one over π₯ by applying some function transformations.
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The graph of the parent function π¦ equals one over π₯ has a horizontal asymptote at π¦ equals zero and a vertical asymptote at π₯ equals zero.
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The given graph has a horizontal asymptote at π¦ equals negative two and a vertical asymptote at π₯ equals three.
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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 π¦ equals one over π₯.
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Before applying this translation, however, we first need to check if any other transformations are involved, since the order of transformations is very important.
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There are three different types of transformations to consider: translation, dilation, and reflection.
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The given graph is oriented the same way as the parent function, so we can rule out reflection.
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Dilation is a possibility, but itβs difficult to judge by eye.
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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.
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Recall that when combining transformations, dilations and reflections must be done before translations.
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Recall that a horizontal dilation by a scale factor of π one means that we map π₯ to π₯ over π one.
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In other words, the π₯-values of all the points on the graph are reduced by a factor of π one.
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A vertical dilation by a scale factor of π two means mapping π of π₯ to π two times π of π₯.
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In other words, we multiply all of the π¦-values on the graph by a factor of π two.
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Beginning with the parent function π of π₯ equals one over π₯ and performing a horizontal dilation, we get one over π₯ over π one.
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Next, performing a vertical dilation, we multiply this new π of π₯ by π two, giving π two over π₯ over π one.
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This simplifies to π one π two over π₯.
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π one times π two is just another constant in and of itself.
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So, we can call this constant π.
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We do not yet know the value of π.
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First, we need to move on to translations.
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The positions of asymptotes are unaffected by dilations.
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Therefore, these could only have been moved by the translation of two units downwards and three units to the right.
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A vertical translation of π units means mapping π of π₯ to π of π₯ plus π.
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So, we change all of the π¦-values of the points on the graph from π¦ to π¦ plus π.
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π being positive means the graph moves upwards and negative means it moves downwards.
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In our case, we need to perform the dilation first.
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So, the parent function one over π₯ goes to π over π₯.
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Then, to translate this by two units downwards, this goes to π over π₯ minus two.
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Now, recall that for a horizontal translation by π units, we map all of the π₯-values to π₯ minus π.
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Again, the value of π will determine the direction, with positive π giving a rightward shift and negative π giving a leftward shift.
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But of course, the sign will be reversed since we are subtracting π from π₯.
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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.
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So, this gives us π over π₯ minus three minus two.
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So, this graph represents the function π of π₯ equals π over π₯ minus three minus two.
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So, we have found two of the parameters in the question.
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π is equal to three and π is equal to negative two.
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Now we just need to find the value of π.
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This equation has three unknowns in it.
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We have π of π₯, π₯, and π.
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We are given a point on the graph: six, negative one.
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And this corresponds to values of π of π₯ and π₯, respectively, which we can then substitute into the equation and rearrange to find π.
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So, we have π of π₯ equals negative one and π₯ equals six.
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Simplifying and rearranging gives π equals three.
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So, we now have the values of all three constants.
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π is equal to three, π is equal to negative two, and π is equal to three.
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In the previous example, we saw how dilation of a graph is difficult to judge by eye.
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So, we used an unknown scale factor, π, and then found this scale factor by using a point on the graph.
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Specifically, if we start with the reciprocal function π¦ equals one over π₯ and perform a horizontal dilation by a scale factor of π one, we get one over π₯ over π one.
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And then performing a vertical dilation by a scale factor of π two gives π two over π₯ over π one, which simplifies to π one π two over π₯.
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Because π one and π two are multiplied together in the end, this transformation makes no distinction between horizontal and vertical dilations.
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For example, if the horizontal dilation factor π one is equal to one and the vertical dilation factor π two is equal to two, then we get two over π₯.
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Conversely, if the horizontal dilation factor is two and the vertical dilation factor is one, we get exactly the same result, two over π₯.
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This is a clear demonstration of the symmetry of the reciprocal function, but it does actually go further than this.
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If we take the function after performing a horizontal and a vertical dilation and reflect it over the π₯-axis, we swap the sign of π of π₯.
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So, this gives negative π one π two over π₯.
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Conversely, if we reflect the function over the π¦-axis, we swap the sign of π₯, which gives π one π two over negative π₯.
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Rearranging this, we get exactly the same result as the reflection over the π₯-axis.
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Whatβs more, just as we did in the previous example, this numerator here is still a constant, which we can call π.
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So, after a horizontal dilation and a vertical dilation and a reflection over either axis or both, we get just π over π₯.
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So, this scalar constant, π, accounts for all horizontal and vertical dilations and all reflections.
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This greatly simplifies the function transformation since we can account for all dilations and reflections by simply multiplying by a constant π.
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Now, all that remains are translations.
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We can perform a horizontal translation by π units by mapping π₯ two π₯ minus π.
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And we vertically translate the graph by π units by adding π to the π¦-values.
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Therefore, a hyperbola with vertical asymptote π₯ equals π and horizontal asymptote π¦ equals π is the graph of a rational function π of π₯ equals π over π₯ minus π plus π, for some π not equal to zero.
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In the next example, we will use this to identify the range of values for an unknown parameter in the given graph of a rational function.
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The graph shows π¦ equals π over π₯ minus three minus two.
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We can see that the intersection of its asymptotes is at three, negative two and that the points 0.5, negative 1.5 and 1.5, negative one are below and above the graph, respectively.
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Determine the interval in which π lies.
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Recall first that a hyperbola with a horizontal asymptote at π₯ equals π and a vertical asymptote π¦ equals π is the graph of the function π of π₯ equals π over π₯ minus π plus π, for π not equal to zero.
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In this case, the vertical asymptote is at π₯ equals three.
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So, π is equal to three.
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And the horizontal asymptote is at π¦ equals negative two.
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So, π is equal to negative two.
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This matches the equation of the given function π of π₯ equals π over π₯ minus three minus two.
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The value of π cannot be determined exactly unless we have any of the points on the graph, which we do not.
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Instead, we have one point above and one point below the graph, which we can use to determine an interval in which π must lie.
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Consider the point above the graph: 1.5, negative one.
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Since this point is above the graph, its π¦-value is greater than π of π₯ at this point.
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In other words, π of 1.5 is less than the π¦-value of this point, negative one.
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Therefore, π over 1.5 minus three minus two is less than negative one.
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Taking the two to the other side and simplifying gives us negative π over 1.5 is less than one.
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And rearranging for π gives us π is greater than negative 1.5.
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We now have a lower bound for π of negative 1.5.
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Now, letβs consider the other point below the graph: 0.5, negative 1.5.
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Since this point is below the graph, its π¦-value is less than the π¦-value of π of π₯ at this point.
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In other words, π of 0.5 is greater than negative 1.5, which means that π over 0.5 minus three minus two is greater than negative 1.5.
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Taking the two over to the right-hand side and simplifying gives us negative π over 2.5 is greater than 0.5.
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Solving for π gives us the upper bound of π.
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π is less than negative 1.25.
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This gives us our final answer, the interval for π.
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π is greater than negative 1.5 and less than negative 1.25.
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Letβs now finish this video by recapping some key points.
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Unlike a graph of a nonconstant polynomial, the graph of a rational function may have vertical and horizontal asymptotes, that is, straight lines which the graph may approach but never touch.
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The graph of π¦ equals one over π₯ is a hyperbola with a horizontal asymptote π¦ equals zero and a vertical asymptote π₯ equals zero.
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And finally, a hyperbola with vertical asymptote π₯ equals π and horizontal asymptote π¦ equals π is the transformed graph of a rational function π of π₯ equals π over π₯ minus π plus π, for some π not equal to zero.