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Alex787 [66]
3 years ago
11

Given that

Mathematics
1 answer:
zloy xaker [14]3 years ago
4 0

g(x) = f(x) + 9 shifts it up 9 units.

g(x) = f(x+2) shifts it left 2 units.

g(x) = -f(x) reflects it across the x-axis.

It looks like you may want

... g(x) = -(f(x+2) +9) . . . . . shifted up, then reflected

g(x)=-\left(5^{(x+2)}+9\right)

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\bf \qquad \qquad \qquad \qquad \textit{function transformations}&#10;\\ \quad \\\\&#10;&#10;\begin{array}{rllll} &#10;% left side templates&#10;f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}&#10;\\ \quad \\&#10;y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}&#10;\\ \quad \\&#10;f(x)=&{{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}&#10;\\ \quad \\&#10;f(x)=&{{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}&#10;\\ \quad \\&#10;f(x)=&{{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}&#10;\end{array}

\bf \begin{array}{llll}&#10;% right side info&#10;\bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\&#10;\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}&#10;\\\\&#10;\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\&#10;\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\&#10;\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\&#10;\end{array}

\bf \begin{array}{llll}&#10;&#10;&#10;\bullet \textit{ vertical shift by }{{  D}}\\&#10;\qquad if\ {{  D}}\textit{ is negative, downwards}\\\\&#10;\qquad if\ {{  D}}\textit{ is positive, upwards}\\\\&#10;\bullet \textit{ period of }\frac{2\pi }{{{  B}}}&#10;\end{array}

now, with that template above in mind, let's see this one

\bf parent\implies f(x)=|x|&#10;\\\\\\&#10;\begin{array}{lllcclll}&#10;f(x)=&3|&1x&+2|&+4\\&#10;&\uparrow &\uparrow &\uparrow &\uparrow \\&#10;&A&B&C&D&#10;\end{array}


A=3, B=1,  shrunk by AB or 3 units, about 1/3
C=2,          horizontal shift by C/B or 2/1 or just 2, to the left
D=4,          vertical shift upwards of 4 units

check the picture below

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