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Find the vertical and horizontal asymptotes of the function π of π₯ equals three π₯ squared minus one over five π₯ squared plus three.
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What is a vertical asymptote of a function?
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Itβs a vertical line that the graph of the function gets closer and closer to.
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And there are various ways that the function can do this.
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The function could approach infinity from the left and negative infinity from the right or vice versa.
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Or both left-hand and right-hand limits could be infinity.
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In which case, the limit itself would be infinity or negative Infinity.
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Or one of the one-sided limits could be finite where the other is infinite.
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Again, there are various ways this could happen for a general function.
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But we have a special kind of function, a rational function.
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And the vertical asymptotes of rational functions come from zeros of the denominator.
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This isnβt to say that every zero of a denominator of a rational function gives a vertical asymptote.
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For example, the rational function π₯ minus one over π₯ minus one has a zero at π₯ equals one.
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But the graph of this function is just the straight line π¦ equals one with one point, the point with π₯-coordinate one removed.
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But the line π₯ equals one isnβt a vertical asymptote of this function.
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So a zero of the denominator doesnβt guarantee an asymptote.
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But if there is an asymptote, it will definitely come from a zero of the denominator.
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What are the zeros of our denominator, five π₯ squared plus three?
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Subtracting three from both sides and dividing by five we see that π₯ squared is a negative number, negative three-fifths.
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And thereβs no real number whose square is negative.
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And so, this equation has no real solutions.
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And so there are no vertical asymptotes.
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We move on to finding any horizontal asymptotes there might be.
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A horizontal asymptote is a horizontal line that the graph of the function approaches.
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There are fewer cases to consider for horizontal asymptotes.
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The line π¦ equals πΏ is a horizontal asymptote of the graph of π of π₯, if either the limit of π of π₯, as π₯ approaches negative infinity, is πΏ or the limit of π of π₯, as π₯ approaches infinity, is πΏ.
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Well, thereβs no limit to the number of vertical asymptotes a functionβs graph could have.
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It can only have two horizontal asymptotes, one as π₯ approaches infinity and the other as π₯ approaches negative infinity.
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To find any horizontal asymptotes therefore, we find the limit of π of π₯ as π₯ approaches negative infinity and the limit of π of π₯ as π₯ approaches infinity.
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We use the definition of π of π₯.
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The trick to evaluating this limit is to divide both numerator and denominator by the highest power of π₯ you can see.
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In this case, π₯ squared.
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After simplifying, we get the limit of three minus one over π₯ squared over five plus three over π₯ squared as π₯ approaches negative infinity.
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And using the fact that the limit of a quotient is the quotient of the limit, as long as the value of the limit in the denominator is nonzero, and also that the limit of a sum or difference is the sum or difference of the limits, we end up with this.
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The limit of the constant function three, and as π₯ approaches negative infinity, is just three.
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And similarly, the limit of the constant function five, as π₯ approaches negative infinity, is just five.
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So we can get rid of these limit signs.
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And the other limits, the limit of one over π₯ squared as π₯ approaches negative infinity and the limit of three over π₯ squared as π₯ approaches negative infinity, are both zero.
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Weβre left with just three over five.
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And so, using our fact about horizontal asymptotes, the line π¦ equals three over five is a horizontal asymptote of the graph of π of π₯.
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That was the limit of πof π₯, as π₯ approaches negative infinity.
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How about the limit of π of π₯, as π₯ approaches just infinity?
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We just replace all the references to negative infinity by positive infinity.
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And we see that we get the same answer, three-fifths.
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So we donβt get a second horizontal asymptote.
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Thereβs only one with equation π¦ equals three-fifths.
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This is therefore our final answer.
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The function has no vertical asymptote and a horizontal asymptote at π¦ equals three-fifths.
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We managed to do this all without graphing the function π of π₯ equals three π₯ squared minus one over five π₯ squared plus three.
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You may now like to graph the function to see that the answer we got is indeed right.