What is the equation for the graph of g(x)?..................edited

eddy2017

eddy2017

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Hi, I will need some help to solve this. Just a pointer to get me in the right direction and I'' take it and run with it.

DIRECTIONS: Read the question and select the best response.
The graph of the equation f(x) = 2xf left paren x right paren is equal to 2 to the power of x is reflected over the x-axis and translated three units to the right to become g(x)g left parenx right paren, as depicted below.
What is the equation for the graph of g(x)g?

  1. g(x) = –2^(x − 3)
  2. g(x) = –2^(x) − 3
  3. g(x) = 2^[–(x − 3)]
  4. g(x) = 2^(–x) − 3
 

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Break it into two steps: first figure out the formula for the reflected version of [imath]f(x)[/imath], then for the shifted version of the reflected one.
 
eddy2017 said:
Hi, I will need some help to solve this. Just a pointer to get me in the right direction and I'' take it and run with it.

DIRECTIONS: Read the question and select the best response.
The graph of the equation f(x) = 2xf left paren x right paren is equal to 2 to the power of x is reflected over the x-axis and translated three units to the right to become g(x)g left parenx right paren, as depicted below.
What is the equation for the graph of g(x)g?

  1. g(x) = –2(x − 3)
  2. g(x) = –2x − 3
  3. g(x) = 2–(x − 3)
  4. g(x) = 2–x − 3
Click to expand...
You need to learn the use of "exponent" sign - which is ^.

I have edited your OP
 
What have you learned about graph transformation? What causes a graph to reflect over the x-axis? What causes a graph to translate up/down or left/right?
 
Subhotosh Khan said:
You need to learn the use of "exponent" sign - which is ^.

I have edited your OP
Click to expand...
please, Dr Khan, can you edit it again.
What is the equation for the graph of g(x)g? wrong
What is the equation for the graph of g(x)?. right. and it is not g raised to the x. it is g(x)
 
When I give a like to your hints means I am gonna be doing my homework on the thing you just pointed out to me. If I take a while is because I am working on it. I will never leave a thread unfinished. Thanks for the tips. I will reflect on what has been said and get back to you.
 
Unless you want to continue to confuse me can you please use a " , " in The graph of the equation f(x) = 2xf left paren x right paren is equal to 2 to the power of x
Please place it where denoted. The graph of the equation f(x) = 2^xf , left paren x right paren is equal to 2 to the power of x.
Otherwise I will think that f(x) = 2^x*f(x) = 2^x. The only reason I understood what you meant is because f(x) = 2^x*f(x) = 2^x makes no sense!
 
Jomo said:
Unless you want to continue to confuse me can you please use a " , " in The graph of the equation f(x) = 2xf left paren x right paren is equal to 2 to the power of x
Please place it where denoted. The graph of the equation f(x) = 2^xf , left paren x right paren is equal to 2 to the power of x.
Otherwise I will think that f(x) = 2^x*f(x) = 2^x. The only reason I understood what you meant is because f(x) = 2^x*f(x) = 2^x makes no sense!
Click to expand...
Sorry. It was when I copied from the exercise page all that carried over .
This is how it should read:
DIRECTIONS: Read the question and select the best response.

The graph of the equation f(x) = 2x is reflected over the x-axis and translated three units to the right to become g(x),as depicted below.

What is the equation for the graph of g(x)?.
 
Hi, I ahve been studying the topic, transformation of graphs, shifts, reflections, etc. It is something new for me. I am still studying it.
But I am going to post one way to answer these question that I learned, of course, it is the easiest for me because I just had to input the equations given in the choice ion Desmos graphing calculator and can clearly be appreciated that choice A is the equation for the graph of g(x).
1642291911758.png
here we can see the reflection over the x-axis moved three units to the right.
 
I will come back again to post the answer without the use of the graphing calculator. The hardest way, of course.
thanks to all
 
Where are you getting the exponentials from?

f(x) = 2x
Reflect f(x) over the y-axis to get -2x.

How do you find g(x) from there?

-Dan
 
Hi, I have been studying some tutorials about this. I am gonna post some work. I signed up for 12 hours of algebra 1 and 2 and Geometry 1 and 2 provided for teachers by our State board so I will be posting less.
the problem gives me the graph of the equation.
I have studied that these transformations of exponential equations may take three forms:
[math]y= b^x[/math][math]y= ab^x[/math][math]y= ab^x-h + k[/math]
According to the choices given to me in the answer choice, Choice A [math]g(X)=-2^(x-3)[/math] is the right choice.
The only think to keep in mind is the number associated to the x value because it is going to be shifting in the opposite direction of the sign this number carries. The number in question is -3 so it will be going to the right 3 units( translated 3 units to the right, as the problem states). This checks out with the graph given in the problem.

Sorry for the delay. I have been studying these in my free time and Algebra and Geometry is taking up the rest of the time and work.
 
The last type of exponential equation is y=ab^x-h +k (just to clear that up)

at post # 9 we can see how it got reflected over the x-axis and moved three units to the right.
f(x) = 2x
g(x) = –2^(x − 3)
the 2 was positive and became negative 2.
f(x) is positive and the g(x) is negative so there is a reflection over the x axis.
the number that accompanies the x in the exponent dictates if the graphs moves to the left or to the right. in this case -3 has the opposite effect on the graph. we tend to think it would go left 3 but it goes right 3.

the only choice that has a negative a base raised to x -a number is A.
The others do not qualifiy.
 
eddy2017 said:
The last type of exponential equation is y=ab^x-h +k (just to clear that up)
Click to expand...
I think you are missing parentheses again. Most probably you wanted to write:

y = a * b^(x - h) + k

Review your post again!!
 
Yes, that's it. In the tutorial it was without parentheses, though.
 
eddy2017 said:
Yes, that's it. In the tutorial it was without parentheses, though.
Click to expand...
If that is truly the case, I suggest use some other tutorial. But I suspect you are interpreting it incorrectly.
 
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