Question

I need help with #11

Answer 1

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

$\bf \begin{array}{llll} % right side info \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\ \bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative} \\\\ \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\ \end{array}$

$\bf \begin{array}{llll} \bullet \textit{ vertical shift by }{{ D}}\\ \qquad if\ {{ D}}\textit{ is negative, downwards}\\\\ \qquad if\ {{ D}}\textit{ is positive, upwards}\\\\ \bullet \textit{ period of }\frac{2\pi }{{{ B}}} \end{array}$

now, with that template in mind, let's take a peek at this function

$\bf \begin{array}{lllcclll} y=&2(&1x&-2)^2&-4\\ &\uparrow &\uparrow &\uparrow &\uparrow \\ &A&B&C&D \end{array}\\\\ -----------------------------\\\\ A\cdot B=2\impliedby \textit{shrunk by a factor of 2, of half-size}\\\\ \cfrac{C}{B}= \cfrac{-2}{1}\implies -2\impliedby \textit{horizontal right shift of 2 units}\\\\ D=-4\impliedby \textit{vertical down shift of 4 units}$

so, the graph of y=2(x-2)²-4, is really the same graph of y=x², BUT, narrower, and moved about horizontally and vertically

Answer 2

11. The 2 in the front shows that the graph is narrowed by a function of 2.
The -2 in the parentheses means it moves to the right 2 units.
The -4 outside the parentheses means it moves down 4 units. 

I have attached the 2 graphs so you can compare.

I hope that helps.