Question

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

Answer 1

`\bf ~~~~~~~~~~~~\textit{function transformations}\\\\\\f(x)= A( Bx+ C)+ D\\\\~~~~y= A( Bx+ C)+ D\\\\f(x)= A√( 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 }( C)/( B)\\~~~~~~if\ ( C)/( B)\textit{ is negative, to the right}`

`\bf ~~~~~~if\ ( 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 }(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.