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yulyashka [42]
3 years ago
13

How does the graph of f(x) =-3^2x-4 differ from the graph of g(x)=-3^2x

Mathematics
1 answer:
Neko [114]3 years ago
3 0
\bf ~~~~~~~~~~~~\textit{function transformations}
\\\\\\
% templates
f(x)={{  A}}({{  B}}x+{{  C}})+{{  D}}
\\\\
~~~~y={{  A}}({{  B}}x+{{  C}})+{{  D}}
\\\\
f(x)={{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}
\\\\
f(x)={{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}
\\\\
f(x)={{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}
\\\\
--------------------

\bf \bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\
~~~~~~\textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{  B}}\textit{ is negative}\\
~~~~~~\textit{reflection over the y-axis}

\bf \bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
~~~~~~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}}\\
~~~~~~if\ {{  D}}\textit{ is negative, downwards}\\\\
~~~~~~if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{  B}}}

with that template in mind, let's check,

\bf f(x)=-3^{\stackrel{B}{2}x\stackrel{C}{-4}}\qquad \qquad g(x)=-3^{2x}\implies  g(x)=-3^{\stackrel{B}{2}x\stackrel{C}{+0}}

notice, g(x) has a horizontal shift of C/B or +0/2, or just 0, none.

while f(x) has a horizontal shift of C/B or -4/2, or -2, to the right.

so f(x) is really just g(x), but shifted horizontally over 2 units to the right.
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Answer:

4 \sqrt{6}  \times  \sqrt{3}  = 12 \sqrt{2}

Step-by-step explanation:

We want to simplify the radical expression:

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We write √6 as √(2*3).

This implies that:

4 \sqrt{6}  \times  \sqrt{3}  = 4 \sqrt{2 \times 3}   \times  \sqrt{3}

We now split the radical for √(2*3) to get:

4 \sqrt{6}  \times  \sqrt{3}  = 4 \sqrt{2}  \times  \sqrt{3}  \times  \sqrt{3}

We obtain a perfect square at the far right.

4 \sqrt{6}  \times  \sqrt{3}  = 4 \sqrt{2}  \times  (\sqrt{3} )^{2}

This simplifies to

4 \sqrt{6}  \times  \sqrt{3}  = 4 \sqrt{2}  \times 3

This gives us:

4 \sqrt{6}  \times  \sqrt{3}  = 4 \times 3 \sqrt{2}

and finally, we have:

4 \sqrt{6}  \times  \sqrt{3}  = 12 \sqrt{2}

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