1/14 - show solution/reason. (5 pts)\narwen studies the two functions shown below.\n$f(x) = 4(2^x)$ $g(x) =…

1/14 - show solution/reason. (5 pts)\narwen studies the two functions shown below.\n$f(x) = 4(2^x)$ $g(x) = 4(2^x) + 5$\nwhat can she say about how the graph of $g(x)$ compares to the graph of $f(x)$?\nthe graph of $g(x)$ is the graph of $f(x)$ shifted 5 units to the right.\nthe graph of $g(x)$ is the graph of $f(x)$ shifted 5 units to the left.\nthe graph of $g(x)$ is the graph of $f(x)$ shifted 5 units up.\nthe graph of $g(x)$ is the graph of $f(x)$ shifted 5 units down.

1/14 - show solution/reason. (5 pts)\narwen studies the two functions shown below.\n$f(x) = 4(2^x)$ $g(x) = 4(2^x) + 5$\nwhat can she say about how the graph of $g(x)$ compares to the graph of $f(x)$?\nthe graph of $g(x)$ is the graph of $f(x)$ shifted 5 units to the right.\nthe graph of $g(x)$ is the graph of $f(x)$ shifted 5 units to the left.\nthe graph of $g(x)$ is the graph of $f(x)$ shifted 5 units up.\nthe graph of $g(x)$ is the graph of $f(x)$ shifted 5 units down.

Answer

Explanation:

Step1: Recall function transformation rules

For a function ( y = f(x) + k ), if ( k>0 ), the graph of ( y = f(x) ) is shifted up by ( k ) units; if ( k < 0 ), it is shifted down by ( |k| ) units. For horizontal shifts, the form is ( y=f(x - h) ), where ( h>0 ) shifts right and ( h < 0 ) shifts left.

Step2: Compare ( f(x) ) and ( g(x) )

Given ( f(x)=4(2^{x}) ) and ( g(x)=4(2^{x})+5 ). We can see that ( g(x)=f(x)+5 ).

Step3: Determine the transformation

Using the vertical shift rule (since we are adding a constant to the entire function ( f(x) )), adding 5 to ( f(x) ) means the graph of ( f(x) ) is shifted up by 5 units to get ( g(x) ). So the graph of ( g(x) ) is the graph of ( f(x) ) shifted 5 units up.

Answer:

The graph of ( g(x) ) is the graph of ( f(x) ) shifted 5 units up. (Corresponding to the option: "The graph of ( g(x) ) is the graph of ( f(x) ) shifted 5 units up.")