Which is a stretch of an exponential decay function? f(x) = 4/5(5/4)^x f(x) =4/5(4/5)^x f(x) =5/4(4/5)^x f(x) =5/4(5/4)^x

Mathematics

2 answers:

ad-work [718]4 years ago

7 0

Answer:

The correct option is 3. The function $f(x)=\frac{5}{4}(\frac{4}{5})^x$ is a stretch of an exponential decay function.

Step-by-step explanation:

The general form of an exponential function is

$f(x)=ab^x$

where, a is the initial value and b is the growth factor.

If b<1, then it represents a decay function and if b>1, then it represents a growth function.

Let k be a stretch and compression factor.

$g(x)=kf(x)$

If k<1, then it represents vertical compression and if k>1, then it represents vertical stretch.

In third function

$f(x)=\frac{5}{4}(\frac{4}{5})^x$

$k>1$

$b$

Therefore the function $f(x)=\frac{5}{4}(\frac{4}{5})^x$ is a stretch of an exponential decay function.

WARRIOR [948]4 years ago

5 0

Answer:choice three

Step-by-step explanation:

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A good doctor recommends that a person who weighs 65 pounds be given 200

Tresset [83]

9514 1404 393

Answer:

295 mg

Step-by-step explanation:

The proportion can be written ...

dose/(96 lb) = (200 mg)/(65 lb)

dose = (96/65)(200 mg) ≈ 295.4 mg . . . . . multiply by 96 lb, cancel lb units

The doctor might recommend 295 mg for a 96-pound patient.

6 0

3 years ago

Write a formula for function f in terms of function g.

emmasim [6.3K]

$\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}$

so hmm if you notice, your f(x) there, looks just like g(x), but is "shifted horizontally" by about 9 units to the right

that simply means, based on that template above, that C/B = -9, and we can simply make B = 1 and C = -9, and we get C/B = -9/1 which is -9

thus $\bf \begin{array}{llcll}
f(x)=g(&1x&-9)\\
&\uparrow &\uparrow \\
&B&C
\end{array}\iff f(x)=g(x-9)$