Question Video: Finding the Second Derivative of a Function Defined by Parametric Equations | Nagwa Question Video: Finding the Second Derivative of a Function Defined by Parametric Equations | Nagwa

# Question Video: Finding the Second Derivative of a Function Defined by Parametric Equations Mathematics • Higher Education

Find the horizontal and vertical asymptotes of π(π₯) = 1/(π₯ β 1) β 2.

02:22

### Video Transcript

Find the horizontal and vertical asymptotes of π of π₯ equals one over π₯ minus one minus two.

We begin by recalling the definition of a horizontal and vertical asymptote. We say that the line π¦ equals πΏ is a horizontal asymptote of a function π¦ equals π of π₯ if the limit as π₯ approaches either negative or positive β of π of π₯ is equal to πΏ. Similarly, π₯ equals π is a vertical asymptote of a function π¦ equals π of π₯ if any of the six statements are true. That is, the limit as π₯ approaches π from the left of the function is either positive or negative β. The limit as π₯ approaches π from the right of the function is either positive or negative β. Or the limit as π₯ approaches π of the function is positive or negative β.

Now, our function π of π₯ is equal to one over π₯ minus one minus two. So weβll begin by evaluating the limit as π₯ approaches β of one over π₯ minus one minus two. And in fact, very little manipulation is required here. As π₯ gets sufficiently large, one over π₯ minus one approaches zero. And so the limit as π₯ approaches β of one over π₯ minus one minus two is zero minus two, which is simply negative two. And in fact, as π₯ approaches negative β, one over π₯ minus one still approaches zero. So the limit as π₯ approaches negative β of π of π₯ is equal to negative two. And we find the equation π¦ equals negative two is a horizontal asymptote of our function.

Next, weβll consider the vertical asymptotes. Less formally, we say that, with rational functions, thatβs those written as the quotient of two functions, vertical asymptotes are the lines which correspond to the zeros of the denominator of the function. So in fact, weβre going to begin by subtracting two from one over π₯ minus one by creating a common denominator. We multiply two over one by π₯ minus one over π₯ minus one. And then, we can subtract the numerator. So we have one minus two times π₯ minus one over π₯ minus one. We simplify. And we find that π of π₯ can be written as three minus two π₯ over π₯ minus one.

Weβre interested in the values of π₯ so that our denominator π₯ minus one is equal to zero. Well, if we solve this equation for π₯ by adding one to both sides, we find that π₯ is equal to one. And so we found the horizontal and vertical asymptotes of π of π₯ equals one over π₯ minus one minus two. The horizontal asymptote is π¦ equals negative two. And the vertical asymptote is π₯ equals one.

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