SOLUTION: How does the graph of the function compare to the parent function? Y=5(2)^x+1+3

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Question 906931: How does the graph of the function compare to the parent function? Y=5(2)^x+1+3
Answer by AlgebraLady88(44) About Me  (Show Source):
You can put this solution on YOUR website!
The equation is y= 5(2)^x+1 +3
The parent function here is y= 2^x
The parent function is just where the graph y= 5(2)^x+1 +3 originates from.
We are talking about transformations of exponential functions here.
Transformations involve translations, reflections, expansions (stretching) or compressions (shrinking) or any combination of these.
Here's what an equation with transformations look like:
y= a(b) ^ c(x-h) + k
Please note:If a is negative, we will flip the graph over the x axis.
a describes either a vertical compression or expansion.
If a> 1, we will have an expansion
If a< 1 , we will have a compression.
b describes exponential growth or decay.
If b>1, we will have growth
If b is bigger than zero but smaller than one , we will have exponential decay
c describes horizontal compression or expansion
If c is bigger than zero but smaller than one , we will have expansion
If c >1 , we will have compression
Note : we do the opposite here
(x-h) is a horizontal shift(translation) left or right.
If the sign in the parenthesis is negative, we will shift right
If the sign in the parenthesis is positive, we will shift left.
Note: Notice we do the opposite here too.
k describes vertical shift (translation) up or down.
If sign in front of k is positive, we move the graph up
If sign in front of k is negative, we move the graph down.
Rules for dealing with a multi transformational exponential growth such as
y= 5(2)^ x+1 +3
1) Do any flipping over the x axis first. This will depend on whether a is negative or not
2) Do the compressions or expansions next( multiplying and dividing)
3) Do the translations ( adding and subtracting)
So, for y= 5(2)^x+1 +3
we see that y= (2)^x has been vertically expanded by a factor of 5, then translated 1 to the left, and then finally, translated 3 up.
There is an asymptote at y=3, so you will find the graph approaching y=3 , but not quite touching it.