Question 1204295: What can you say about the functions graphed below? Explain your reasoning: Is it a log or an exponential? For an exponential, is the exponent x or -x? What is the equation for the asymptote? Which direction and how many units is it shifted?
Graph A) https://app.gemoo.com/share/image-annotation/572889363786969088?codeId=Ml2kEYAjRgoXb&origin=imageurlgenerator
Graph B) https://app.gemoo.com/share/image-annotation/572889769845932032?codeId=v6gK8ZKQR0YZa&origin=imageurlgenerator
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
Graph A represents a log function.
This particular curve passes through (3,3) and (12,6).
Consider the log equation template
y = a + b*log(x)
If we plugged in x = 1, then we get y = a.
This is because log(1) = 0 regardless of the base.
Unfortunately x = 1 isn't defined on your log graph.
But let's shift (3,3) and (12,6) two units to the left so that x = 1 would be defined.
(3,3) moves to (1,3)
(12,6) moves to (10,6)
If we plugged the coordinates of (1,3) into y = a+b*log(x), then we'd find that a = 3.
So y = a+b*log(x) updates to y = 3+b*log(x)
Now try the coordinates of (10,6) to determine b.
y = 3+b*log(x)
6 = 3+b*log(10)
6 = 3+b*1 ...... I'm using log base 10
6 = 3+b
b = 6-3
b = 3
We have y = 3 + b*log(x) update to y = 3 + 3*log(x)
This would be the answer if (1,3) and (10,6) were points on the curve.
However, recall we shifted things over 2 units to the left.
Shift 2 units to the right to undo that previous shift.
This means we'll replace x with x-2 to apply this shift we need.
The function that represents graph A is y = 3 + 3*log(x-2) where the log is base 10.
The vertical asymptote is x = 2 because this x value makes the stuff inside the log, the x-2 portion, equal to zero.
Recall that log(x) has the domain x > 0
Therefore the domain of y = 3 + 3*log(x-2) is x > 2.
The curve looks like it touches x = 2, but it never actually gets there. Think of it like an electric fence.
-------------------------------------------------------------------------------------------------------------------------
Now onto graph B. It appears this curve goes through (1,8) and (3,2).
This is an exponential decay curve.
This is due to the fact it decreases as we move to the right.
A real world example would be a half-life function.
The exponent is -x to indicate this decay.
The horizontal asymptote of the function y = a*b^(-x) is y = 0.
This curve shows the horizontal asymptote is y = 1.
Let's shift everything down by 1 so the horizontal asymptote y = 1 overlaps with the x axis (aka y = 0).
(1,8) moves to (1,7)
(3,2) moves to (3,1)
Let's look at the template
y = a*b^(-x)
To make the b term go away, we would plug in x = 0. But it's not clear what x = 0 leads to based on the graph you've provided (even after the vertical shift downward).
Let's move those points over 1 spot to the left.
(1,7) moves to (0,7)
(3,1) moves to (2,1)
Plug in the coordinates of (0,7) to find y = a*b^(-x) leads to a = 7.
So we now have y = 7*b^(-x)
Plug in (2,1) and solve for b
y = 7*b^(-x)
1 = 7*b^(-2)
1 = 7/(b^2)
1*b^2 = 7
b^2 = 7
b = sqrt(7)
b = 7^(1/2)
b = 7^(0.5)
So,
y = 7*b^(-x)
y = 7*(7^(0.5))^(-x)
y = 7*7^(-0.5x)
Next we'll shift 1 unit up and 1 unit right to undo those previous two shifts we did earlier.
y = 7*7^(-0.5x)
becomes
y = 7*7^(-0.5(x-1))+1
I'll let you simplify if needed.
There is probably a much more elegant, clever, and efficient way to determine this function.
I'll let another tutor provide that route. Feel free to explore other methods.
I recommend using a tool like GeoGebra or Desmos to confirm we have the correct functions.
Here is a link to the interactive Desmos graph
https://www.desmos.com/calculator/wc4mvapjx7
|
|
|
