WEBVTT
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the point two negative one called P lies on the
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curve. Michaels 1/106. We know that that is
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true because if we put x equals two in this
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equation here we get Y equals-1. So the
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point to negative one less than the curve. Why
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? It was 1/1-6 in part A. If
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we take any point of that care Q. That
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is with coordinates X. and one over one minute
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six. We want to calculate the slope of the
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second line peak. You for the following values of
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X 1.519199. 2.5 to bring 1 2.01 2.001.
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That is we're taking values closer and close to two
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from the left and from the right. Okay.
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In part cheerier use we're gonna use the results in
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part A. And with that we're going to get
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the value of the slope of the tangent line to
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the cure at B 2-1. You know these
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slow will be the values that are approaching those slopes
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of the second lines you perceive? We use the
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slow guest in barbie To find an equation to turn
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to the lines of the cure at-2-1.
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So we're going to calculate them for a all the
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values. So first we know it. You're following
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. All right, mm hmm. This lobe of
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this second nine through the point p 82 and 81
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and Q Given by X 1/1-X. Supposing X
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is not zero. Sorry X is not one which
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is the value that notifies dominator and the value is
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undefined. So if we write that slope of that
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second line we get Mhm The following it's called the
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globe M. And we know the equation will be
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won over one minus X. Which is the Y
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coordinate of the point Q minus. Do I coordinate
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of the point P. That is negative one.
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My understand the one and in the denominator yet the
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first coordinative Q. That is X minus the first
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coordinator of P. That is to And this give
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us one over 1-6 plus one over eggs minus
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two. And in fact this is equal to one
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minus eggs as denominator in the numerator and um one
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minute six denominator And where do we get one plus
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one minus x. Which is two minus six.
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The denominator of the big fraction we get x minus
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two and then this is equal to two minus x
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over the product of one minus x. Times explain
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this to. We noticed that in the numerator we
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can get I can factor negative one out and then
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this negative parenthesis X-2 over. Okay 1-6
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times X-2. And because X is if we
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take X different front to for example and in this
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case we verify we are not taking the value to
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we are close enough to to where we are never
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taking the value to. So we can say that
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this equal for X. Different from two. That
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we can simplify the factor X-2. And we
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get 91 over one minus six. Or in fact
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one over X-1. So the slope of the
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second line, mhm BQ is m equals one over
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x minus or so. This is our expression here
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. And now we can calculate all these slopes for
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x equal 1.5 which is the first value here we
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get then the slope one over 1.5 minus one is
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1/0 10.5. Remember 0.51 house. So I want
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to now far too. We take x equal 19
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. And for 1.9 we get em equal 1/1 9
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minus one which is 1/0 10.9. And here we
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use the calculator and we find that this is approximately
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equal to and we are going to use six testicle
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figures. So at one one in all the vegetables
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. So this is uh the slope of the second
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line peak you at x equals 1.9. Now,
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Part three. We take 1.99. Yeah. Yeah
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. Okay. And so the slope of the 10
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. 2nd line is one over 199-1 which is
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one over 0.99. And is this approximately equal to
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on 21? Yeah 0101. Then our four x
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is 1.999. And then we get em equal 1/1
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0.999 minus one which is 1/0 10.999. And that's
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approximately equal to 1.001 your little one. No Part
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five. We use x equal We did this already
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. We take 2.5. and so m is one
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over 2.5-1, sequel to one over 1.5.
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And this is approximately equal to some 4.6666 67.
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Mhm. Then for six x is taken equal,
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2.1 Mm is one over 2.11 is one is 1/1
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10.1. That's approximately equal to 09 zero 90 90
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, not zero because the following pressing over the announcer
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, we won here and for seven is equal to
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2.01. and so um is one over 2.01-1
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, which is one over 1.01 and that's approximately two
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. It's your point 99 zero 99 mm Yes.
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Sarah, So this is seven now eight exist.
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2.001. So um is 1/2 0.1 minus one,
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which is one over 1.001. That's approximately equal to
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to your point three times nine 001. Okay,
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yeah. Okay, so these are approximations. So
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we see that yes, first approximations of X closer
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to the value to hear. But from the left
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is all this result up to here and for values
100
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of X Closer to two. But from left,
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the value of the slope of the second line peak
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, you are getting Closer and closer to one with
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values greater than one. Now the other part here
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is for ex closer to two. But from the
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right and for those values of X closer to two
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from the right again develops of the slope of the
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second line peak. You are closer to one again
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but now from values less than But in both cases
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we see that the slope of the second line peak
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, you are getting closer and closer to one so
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far. B we can say that mhm. The
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slope of the tangent line which is in fact the
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limits of this or the values that are approaching these
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slopes of second lines. When we are close to
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be acceptable to slope of the tangent line to the
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cure at P 2-2. Because here we are
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, we have the first question equal to which is
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the value. We are approaching for the variable X
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. And when we approached the X22 from the left
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from the right, the values of the slope of
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the second lines Approaches of only one. And so
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we can say that the slope of the tangent line
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which is defined as those mm limits Which got to
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be equal in fact from the left and from the
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right. And then we say that that slope of
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the tangent line to the care at that point to
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an 81 is one. And in party we know
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then that the line is tangent line has an equation
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of the tangent. Flying to the curve at B291
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is given by Why-2. Using the coordinates of
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the point equal to slow. Calculated by using the
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slopes of the second lines. When x is approaching
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the value to one times X minus two. So
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we get y plus two, Y Plus one.
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Sorry, Equal X-2. And if we put
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all from to the right of the equation we get
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X-Y-3 equals zero. So this is equation
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of the tangent line to the curve At P 2
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91. Yeah. And we have calculated the slope
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of the tangent line to the curve at that point
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. B tune 81. Using the slopes of the
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second lines to the cure fixing the point P two
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negative one and moving the point Q. By changing
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its first coordinate X. And putting it closer and
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closer to two from the writing from the left.
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And in both cases we have seen that the slopes
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of that of those second lines approaches the value one
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. So the slope of the tangent line will be
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one. And with that we have calculated the equation
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of the tangent line at the point two negative one
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--> A:middle L:90%
.