Adi3
  • Adi3
Will medal Please help. Sketch the following functions on the same set of axes as y = x^2. In each case, state the coordinates of the vertex. y = -(x-1)^2 + 3
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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Adi3
  • Adi3
@Jhannybean
Adi3
  • Adi3
@dan815
Adi3
  • Adi3

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Adi3
  • Adi3
@CallMeKiki
Adi3
  • Adi3
@Camila1315
Adi3
  • Adi3
@Miracrown
Adi3
  • Adi3
@IrishBoy123
IrishBoy123
  • IrishBoy123
|dw:1443858498605:dw|
IrishBoy123
  • IrishBoy123
|dw:1443858518127:dw|
IrishBoy123
  • IrishBoy123
ie take it step by step
Adi3
  • Adi3
i didn't get that
Adi3
  • Adi3
IrishBoy123
  • IrishBoy123
ok late's start with \(y = x^2\) what does \(y = (x-1)^2\) look like?
Adi3
  • Adi3
(x-1)(x-1), can't Foil it??
IrishBoy123
  • IrishBoy123
no need for that whereas y = 0 when x = 0 for \(y = x^2\), y = 0 when x = 1 for \(y = (x-1)^2\) it's a shift of the curve, yes?
Adi3
  • Adi3
so it moves one to the right.
IrishBoy123
  • IrishBoy123
indeed!
IrishBoy123
  • IrishBoy123
now let's compare \(y = x^2 \) with \(y = x^2 +3\) that's just another shift
Adi3
  • Adi3
so it moves 3 point up.
IrishBoy123
  • IrishBoy123
indeed now what about \(y = x^2\) and \(y = -x^2\) ??
Adi3
  • Adi3
reflect horizontally????
IrishBoy123
  • IrishBoy123
yes, you can see it that way. a reflection in the horizontal axis or a 180 rotation. either is good |dw:1443859072630:dw|
IrishBoy123
  • IrishBoy123
in your case you have all three at once, ie the horizontal and vertical shifts and the reflection \(y = x^2 \rightarrow y = -(x-1)^2 + 3\)
IrishBoy123
  • IrishBoy123
hence |dw:1443859195875:dw|
Adi3
  • Adi3
thank you. a lot. really needed help!!!!!!!!, thank you mate.
IrishBoy123
  • IrishBoy123
mp!

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