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Aaron Thomas Jacob Wefel Tyler Sneen By the end of this lesson we will introduce the terminology that is used to describe functions These include: Domain, Range, Continuity, Discontinuity, upper and lower bound, Local and absolute maximums and minimums, and asymptotes The domain of a function is all of the possible x-values the function can have. It can be expressed as an inequality The Range of a function is all of the possible y-values the function can have. It is also expressed as an inequality Domain: All Real Numbers Range: All Real Numbers Domain: x> -1 Range: x>-5 A graph has continuity if its graph is connected to itself throughout infinity. There are no asymptotes or holes in the graph A Graph has removable discontinuity if its graph has a hole where one x value was removed from the domain A graph has infinite discontinuity if its graph has an asymptote that can not be replaced with only one value Jump Discontinuity Removable Discontinuity A function is bounded above or below if the graph’s range doesn’t extend past a certain point above or below. A function is “Bounded” if the function’s range doesn’t extend below or above certain points If the function has no restrictions on its range’s extent the function is considered “unbounded” This sine function is bounded above and below at 1 and -1 A Local Maximum/Minimum of a function is the highest/lowest point of the range in the surrounding window of the graph The absolute maximum/minimum of a function is the highest/lowest point of the entire range of the graph Local Min: 3, -4, 4 Local Max: 5 Absolute Max: None (Graph goes infinitely upward) Absolute Min: -4 A horizontal asymptote is a part of the function which gets infinitely close to a Yvalue but never touches it A Vertical asymptote is a part of the function which gets infinitely close to a x- value but never touches it Identify any horizontal or vertical asymptotes of the graph of You would first start by foiling the denominator… = (x+1)(x-2) This means that the graph has vertical asymptotes of x=-1 and x=2 Because the denominator’s power is bigger than the numerator’s, y = 0 no matter what the value of x is Now you have x/((x+1)(x-2)) = 0 This means that the horizontal asymptote is zero