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(a) Find the $x$ -intercept(s); (b) Find the vertical asymptotes; (c) Find the horizontal asymptotes. (d) Sketch the graph.$$f(x)=(2-x)(x+1)$$

(a) $x=2$(b) $x=-1$(c) $y=-1$(d)

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 4

Limits at Infinity, Infinite Limits and Asymptotes

Derivatives

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04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

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In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

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(a) Find the $x$ -intercep…

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(a) Determine the $x$ -int…

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For the graph of $y=f(x)$<…

here were given the function F of X is equal to two minus X, divided by X plus one. And for part A we want to find our X intercept and then we're gonna work through parts B and C. And then hopefully be able to sketch a graph of it at the end. So finding these X intercepts, we just need to set this function equal to zero and solve for X. So two minus X, divided by X plus one. Call zero, multiplying both sides by X plus one. So we can get rid of that fraction. We have two minus X is equal to zero and then let's add X to both sides. And we can see that two is equal to X or X is equal to two and that is going to be our X intercepts. We can go ahead and plot that real quick. So when X is equal to two, well, have a point there, we know our graph is going to intercept it at some point and then we have part B. Finding these vertical ascent oats. And what's important with vertical ascent owes is to look at our denominator. So X plus one really, we just need to take that denominator set it equal to zero and solve for X. You can see here that X is equal to negative one. So at that point X is equal to negative one. That's giving us that vertical ass until it so that's gonna sit right in here. Alright, part C. We want to find this horizontal ass in total. For here we want to pay attention to those dominant terms in the numerator and denominator. So in the numerator that's going to be negative X. Denominator is positive X. We just need to simplify this so you can see that these two X's cancel out and we're just still left with that negative. So a negative one, meaning that y is equal to negative one. That gives us that horizontal as until so we can go ahead and sketch that as well, right in here. So this is giving us kind of these guidelines or a relative outline of what our graph is going to look like. We're not going to have a graph cross any of these lines. We don't quite know just with the information we have where any turning point is going to occur. So go ahead and graph it and your graphing calculator. So you can get a better image of it. But as you can see, the graph does follow our general outline that we just gave ourselves and it looks something like that, right? It is going through our intercept at X equals two and it is staying within those a sum total of X is equal to negative one. And why is also equal to negative one.

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