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slavikrds [6]
2 years ago
13

Explain how the graph of y=|x+1|-2 will differ from the graph of y=|x|.

Mathematics
1 answer:
Natali5045456 [20]2 years ago
3 0

\bf ~~~~~~~~~~~~\textit{function transformations}\\\\\\f(x)=  A(  Bx+  C)+  D\\\\~~~~y=  A(  Bx+  C)+  D\\\\f(x)=  A\sqrt{  Bx+  C}+  D\\\\f(x)=  A(\mathbb{R})^{  Bx+  C}+  D\\\\f(x)=  A sin\left( B x+  C  \right)+  D\\\\--------------------\\\\\bullet \textit{ stretches or shrinks horizontally by  }   A\cdot   B\\\\\bullet \textit{ flips it upside-down if }  A\textit{ is negative}

\bf ~~~~~~\textit{reflection over the x-axis}\\\\\bullet \textit{ flips it sideways if }  B\textit{ is negative}\\~~~~~~\textit{reflection over the y-axis}\\\\\bullet \textit{ horizontal shift by }\frac{  C}{  B}\\~~~~~~if\ \frac{  C}{  B}\textit{ is negative, to the right}

\bf ~~~~~~if\ \frac{  C}{  B}\textit{ is positive, to the left}\\\\\bullet \textit{ vertical shift by }  D\\~~~~~~if\   D\textit{ is negative, downwards}\\\\~~~~~~if\   D\textit{ is positive, upwards}\\\\\bullet \textit{ period of }\frac{2\pi }{  B}

with that template in mind,

\bf y=\stackrel{A}{1}|\stackrel{B}{1}x\stackrel{C}{+1}|\stackrel{D}{-2}

so is really just |x| but shifted some,

C/B = 1/1 = +1, horizontally to the right by 1 unit.

D = -2, vertically downwards by 2 units.

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{ \boxed{{ \bold{{ \pink{e}}{ \blue{x}}{ \purple{p}}{ \green{l}}{ \red{a}}n{ \purple{a}}{ \orange{t}}{ \red{i}}on}}}}

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<h3>Answer Details</h3>
  • Lesson : Math
  • Grade : 4

#I hope this helps°^°

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