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Starting with the graph of y = e<sup>x</sup>, describe an equivalent transformation and find the equation of the graph from the following changes. (a) reflecting about the line y = 8 This is equivalent to reflecting across the ---Select--- and then translating 16 units ---Select--- y = (b) reflecting about the line x = 4 This is equivalent to reflecting across the ---Select--- and then translating 8 units ---Select--- y =

          Starting with the graph of y = e<sup>x</sup>, describe an equivalent transformation and find the equation of the graph from the following changes.
(a) reflecting about the line y = 8
This is equivalent to reflecting across the ---Select--- and then translating 16 units ---Select---
y =
(b) reflecting about the line x = 4
This is equivalent to reflecting across the ---Select--- and then translating 8 units ---Select---
y =
        
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Starting with the graph of y = e<sup>x</sup>, describe an equivalent transformation and find the equation of the graph from the following changes.
(a) reflecting about the line y = 8
This is equivalent to reflecting across the —Select— and then translating 16 units —Select—
y =
(b) reflecting about the line x = 4
This is equivalent to reflecting across the —Select— and then translating 8 units —Select—
y =

Added by Michael S.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Starting with the graph of y = e^(x), describe an equivalent transformation and find the equation of the graph from the following changes. (a) reflecting about the line y = 8 This is equivalent to reflecting across the x-axis and then translating 16 units up y = e^(-x) + 8 (b) reflecting about the line x = 4 This is equivalent to reflecting across the y-axis and then translating 8 units to the left y = e^(8 - x)
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Transcript

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00:01 The way i'm seeing this problem is, so the graph of e to the x, that's poorly drawn, but that's okay, looks something like this.
00:10 So i'll just write that out.
00:13 But then what happens is we're talking about this horizontal line, y equals eight.
00:18 So if we want to reflect over that y equals eight, we need to figure out basically this kind of formula.
00:27 And what i'm thinking about is where that point of intersection would be.
00:33 Well, let me put it this way, is if this ordered pair is zero, one, i drew that very poorly now that i'm thinking about it.
00:47 That graph would actually look something like this.
00:51 There we go, that's much better.
00:54 Okay, so that zero, one would be seven units away from that line we're reflecting over.
01:01 So we need to be seven units above...
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