Question

Disregard my work, but how do you get the answer? There is also a graph at the bottom which is optional to use.

Answer 1

`\bf ~~~~~~~~~~~~\textit{function transformations} \\\\\\ % templates 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 \\\\ --------------------`


`\bf \bullet \textit{ stretches or shrinks horizontally by } A\cdot B\\\\ \bullet \textit{ flips it upside-down if } A\textit{ is negative}\\ ~~~~~~\textit{reflection over the x-axis} \\\\ \bullet \textit{ flips it sideways if } B\textit{ is negative}\\ ~~~~~~\textit{reflection over the y-axis}`


`\bf \bullet \textit{ horizontal shift by }( C)/( B)\\ ~~~~~~if\ ( C)/( B)\textit{ is negative, to the right}\\\\ ~~~~~~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.

since the original or "parent" function is y = x²−4x+3, if we change the "x" argument to "x-2", we end up with a horizontal shift.

the x-2 part would be in the template the Bx+C part, with B = 1 and C = -2, or a horizontal shift to the right of 2/1 or 2 units.

since the parent function has a point at (2, -1), if we move that horizontally only to the right, we'd end up at (4, -1).