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date:
class period
transformations learning activity
lt28: i can identify the transformations of an exponential equation from the parent function.
1. go to desmos. type in the function ( f(x)=2^{x} ). this is your original function. then below it type in the function ( g(x)=2^{x}+5 ). this is your transformed function. use the two graphs that appear to answer the questions below.
what was the asymptote of ( f(x) ) ? what was the asymptote of ( g(x) ) ?
how did the function move?
2. adjust desmos. leave in the function ( f(x)=2^{x} ). this is your original function. then below it type in the function ( g(x)=2^{x}-3 ). this is your transformed function. use the two graphs that appear to answer the questions below.
what was the asymptote of ( f(x) ) ? what was the asymptote of ( g(x) ) ?
how did the function move?
3. adjust desmos. leave in the function ( f(x)=2^{x} ). this is your original function. then below it type in the function ( g(x)=2^{x - 4} ). this is your transformed function. use the two graphs that appear to answer the questions below.
what was the asymptote of ( f(x) ) ? what was the asymptote of ( g(x) ) ?
how did the function move?
4. adjust desmos. leave in the function ( f(x)=2^{x} ). this is your original function. then below it type in the function ( g(x)=2^{x + 4} ). this is your transformed function. use the two graphs that appear to answer the questions below.
what was the domain of ( f(x) ) ? what was the domain of ( g(x) ) ?

Question

Accepted Answer

# Explanation:
## Step1: Analyze the parent function $ f(x) = 2^x $
For an exponential function $ y = a^x$ ($a>0,a
eq1$), the horizontal asymptote is $y = 0$, and the domain is all real numbers, $x\in(-\infty,\infty)$.

## Step2: Analyze $g(x)=2^x + 5$
For a function $y = f(x)+k$, when $k = 5>0$, it is a vertical shift. The horizontal asymptote of $y = 2^x+5$ is $y=0 + 5=5$. The function $g(x)$ is the graph of $f(x)$ shifted up by 5 units.

## Step3: Analyze $g(x)=2^x - 3$
For a function $y = f(x)+k$, when $k=- 3<0$,  the horizontal asymptote of $y = 2^x-3$ is $y=0-3=-3$. The function $g(x)$ is the graph of $f(x)$ shifted down by 3 units.

## Step4: Analyze $g(x)=2^{x - 4}$
For a function $y = f(x - h)$, when $h = 4>0$,  the horizontal asymptote of $y=2^{x - 4}$ remains $y = 0$ (because horizontal shifts do not affect the horizontal asymptote of an exponential function). The function $g(x)$ is the graph of $f(x)$ shifted right by 4 units.

## Step5: Analyze $g(x)=2^{x + 4}$
For a function $y = f(x+h)$ ($h = 4>0$), the domain of an exponential function $y=a^{x + h}$ ($a>0,a
eq1$) is still all real numbers. Because for any real - valued $x$, the expression $x + 4$ is a real number, and $a^{x+h}$ is well - defined. So the domain of $g(x)=2^{x + 4}$ is $x\in(-\infty,\infty)$

# Answer:
1.
- Asymptote of $f(x)$: $y = 0$
- Asymptote of $g(x)$: $y = 5$
- Movement: Shifted up 5 units
2.
- Asymptote of $f(x)$: $y = 0$
- Asymptote of $g(x)$: $y=-3$
- Movement: Shifted down 3 units
3.
- Asymptote of $f(x)$: $y = 0$
- Asymptote of $g(x)$: $y = 0$
- Movement: Shifted right 4 units
4.
- Domain of $f(x)$: $(-\infty,\infty)$
- Domain of $g(x)$: $(-\infty,\infty)$

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QUESTION IMAGE

Question

Explanation:

Step1: Analyze the parent function \( f(x) = 2^x \)

Step2: Analyze \(g(x)=2^x + 5\)

Step3: Analyze \(g(x)=2^x - 3\)

Step4: Analyze \(g(x)=2^{x - 4}\)

Step5: Analyze \(g(x)=2^{x + 4}\)

Answer: