Question

Describe how the graph of y=
`x^(2)` can be transformed to the graph of the given equation.

y =
`(x-12) ^(2)` + 3

Answer 1

`\bf \qquad \qquad \qquad \qquad \textit{function transformations} \\ \quad \\\\ % left side templates \begin{array}{llll} f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}} \\ \quad \\ y=&{{ A}}({{ B}}x+{{ C}})+{{ D}} \\ \quad \\ f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}} \\ \quad \\ f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}} \\ \quad \\ f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}} \end{array}\\\\ --------------------`


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


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

with that template in mind, let's see


`\bf y=(x-12)^2+3\implies y=\stackrel{A}{1}(\stackrel{B}{1}x\stackrel{C}{-12})^2\stackrel{D}{+3}`

A = 1, so no change there from the parent.

B = 1, no change there either

C = -12, now, C/B is -12/1 or -12, that means a horizontal right shift of 12

D = +3, this means, going up by 3 units.