155k views
1 vote
How does the graph of y=-(2x+6)^2+3 compare to the graph of the parent function?

User Mweerden
by
8.0k points

1 Answer

0 votes

\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}}}

now, with that template in mind, let's see,


\bf y=\stackrel{parent~function}{x^2}\implies y=\stackrel{A}{1}(\stackrel{B}{1}x\stackrel{C}{+0})^2\stackrel{D}{+0} \\\\\\ y=\stackrel{A}{-1}(\stackrel{B}{2}x\stackrel{C}{+6})^2\stackrel{D}{+3}\quad \begin{cases} A=-1&\textit{reflection over x-axis}\\ B=2\\ C=6&\textit{horizontal shift of }(C)/(B)\to (6)/(2)\to 3\\ &\textit{to the left}\\ D=3&\textit{vertical shift upwards of 3 units} \end{cases}
User Bostjan
by
8.1k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories