Question

I need help with #11

Answer 1

`\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.