USE the RULE (x,y) -----> (x - 2, y - 4) to graph the image of this rectangle.

1 answer:

3 0

So for example B(4,1) so x=4, y=1
->B'(4-2,1-4)
->B'(2,-3)

->A'(1-2,1-4)
->A'(-1,-3)

->C'(4-1,-2-4)
->C'(3,-6)

->D'(1-2,-2-4)
->D'(-1,-6)

so the image of the rectangle ABCD->A'B'C'D'

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F(x) = x2 – 3x – 2 is shifted 4 units left. The result is g(x). What is g(x)?

Ray Of Light [21]

$\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... your expression is not in vertex form, that's ok... to do a horizontal shift to the left by 4 units, we can simply, add the C and B component to the "x" variable, C=4, B =1, that way the horizontal shift of C/B or 4/1 is just +4, giving us a horizontal shift to the left

$\bf f(x)=x^2-3x-2\impliedby \textit{let's change that for }f(1x+4)
\\\\\\
f(1x+4)=(1x+4)^2-3(1x+4)-2
\\\\\\
f(x+4)=x^2+8x+16-3x-12-2
\\\\\\
f(x+4)=x^2-5x+2\impliedby g(x)$

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Eddi Din [679]

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