What Happens to the Exponential Function y = e^(ax) When 'a' Changes?

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The discussion focuses on the effects of the constant 'a' in the exponential function y = e^(ax). When 'a' is positive, the curve rises steeply, while a negative 'a' causes it to decline. As 'a' increases, the value of y grows significantly faster than x. The participants also explore the significance of the constant 'e', noting its unique properties, including the tangent's angle at y = 1 being 45 degrees. Overall, the conversation emphasizes understanding how changes in 'a' influence the graph's shape and behavior.
k.udhay
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Hi,
In the equation " y = e^(ax), what is the effect of constant 'a'. Like, what happens to the shape of the curve when it becomes -, +, high or low? Thanks.
 
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hi k.udhay! :wink:

(try using the X2 button just above the Reply box :wink:)

tell us what you think, and then we'll comment! :smile:
 
Hi tiny-tim,

I don't see any X^2 button... Thanks.
 
k.udhay said:
I don't see any X^2 button... Thanks.

you'll see it if you click the "Quote" button or the "Go Advanced" button :smile:
 
Hi tiny-tim,

About X2, understood the point. :)
Well, what I can derive easily is when x turns 0, the curve is a straight vertical line. For '+' x, it lies on right hand side, for '-' side the curve travels in left hand side.

Ah... Now, higher the 'x' value, more it will become flat towards right... Am I right?
 
k.udhay said:
Well, what I can derive easily is when x turns 0, the curve is a straight vertical line.

uhh? :confused:

which way up are you? :biggrin:
 
I am really sorry... It should be a horizontal line having an 'Y' interception 1... Correct now?
 
Well let's take x=1.

for a=1, y=e^1
a=2, y=e^2
a=3, y=e^3

So what is happening to the value of y as 'a' increases?
 
Y increases much faster than x... :)
 
  • #10
k.udhay said:
Y increases much faster than x... :)

So can you now see the effect on the graph if 'a' is increased or decreased?
 
  • #11
Assuming you have the graph of y = f(x), here are a few variants.
  • The graph of y = f(x) + a is a vertical translation (or shift) of the graph of y = f(x). If a > 0, the shift is upward. If a < 0, the shift is downward.
  • The graph of y = f(x - a) is a horizontal translation (or shift) of the graph of y = f(x). If a > 0, the shift is to the right. If a < 0, the shift is to the left. For example, the graph of y = (x - 2)2 looks like the graph of y = x2, but shifted horizontally to the right. Instead of the vertex being at (0, 0), the vertex in the shifted graph is at (2, 0).
  • The graph of y = -f(x) is a reflection across the horizontal axis of the graph of y = f(x).
  • The graph of y = f(-x) is a reflection across the vertical axis of the graph of y = f(x). For example, the graph of y = sin(-x) looks like the graph of y = sin(x), but reflected across the y axis.
  • The graph of y = af(x) represents an expansion away from the horizontal axis if a > 1, and a compression toward the horizontal axis if 0 < a < 1. If a < 0, there is also a reflection across the x-axis.
  • The graph of y = f(ax) represents a compression toward the vertical axis if a > 1, and a compression away from the vertical axis if 0 < a < 1. If a < 0, there is also a reflection across the y-axis.
 
  • #12
rock.freak667 said:
So can you now see the effect on the graph if 'a' is increased or decreased?

Thanks rock.freak. Yeah, I think I can figure out the other cases. Now, let me ask my second question:
What is the need for finding this term 'e' which has a strange value of 2.718... When I read in wiki., it says a curve following ex will have its tangent at Y = 1 at an angle of 45°. But why does one need this combination?:confused:
 
  • #13
Mark44 said:
Assuming you have the graph of y = f(x), here are a few variants.
  • The graph of y = f(x) + a is a vertical translation (or shift) of the graph of y = f(x). If a > 0, the shift is upward. If a < 0, the shift is downward.
  • The graph of y = f(x - a) is a horizontal translation (or shift) of the graph of y = f(x). If a > 0, the shift is to the right. If a < 0, the shift is to the left. For example, the graph of y = (x - 2)2 looks like the graph of y = x2, but shifted horizontally to the right. Instead of the vertex being at (0, 0), the vertex in the shifted graph is at (2, 0).
  • The graph of y = -f(x) is a reflection across the horizontal axis of the graph of y = f(x).
  • The graph of y = f(-x) is a reflection across the vertical axis of the graph of y = f(x). For example, the graph of y = sin(-x) looks like the graph of y = sin(x), but reflected across the y axis.
  • The graph of y = af(x) represents an expansion away from the horizontal axis if a > 1, and a compression toward the horizontal axis if 0 < a < 1. If a < 0, there is also a reflection across the x-axis.
  • The graph of y = f(ax) represents a compression toward the vertical axis if a > 1, and a compression away from the vertical axis if 0 < a < 1. If a < 0, there is also a reflection across the y-axis.

Thank you, mark. Your exlanation will help me extimate the behaviour of curves very easily!:approve:
 
  • #14
k.udhay said:
Thanks rock.freak. Yeah, I think I can figure out the other cases. Now, let me ask my second question:
What is the need for finding this term 'e' which has a strange value of 2.718... When I read in wiki., it says a curve following ex will have its tangent at Y = 1 at an angle of 45°. But why does one need this combination?:confused:

You can consider 'e' to just be another special constant without having to go into the details of how to define it. Much like π.

For the y=1, θ=45° thing, you don't really have to memorize something like that but you can derive that as follows:

y=ex → dy/dx = ex i.e. dy/dx =y

So that when y=1, dy/dx =1 i.e. the gradient of the tangent at y=1 is 1.

The angle a straight line makes with the x-axis of gradient 'm' is given by tanθ=m or θ=tan-1(m) so in this case, m=1 such that θ=45°
 
  • #15
rock.freak667 said:
You can consider 'e' to just be another special constant without having to go into the details of how to define it. Much like π.

For the y=1, θ=45° thing, you don't really have to memorize something like that but you can derive that as follows:

y=ex → dy/dx = ex i.e. dy/dx =y

So that when y=1, dy/dx =1 i.e. the gradient of the tangent at y=1 is 1.

The angle a straight line makes with the x-axis of gradient 'm' is given by tanθ=m or θ=tan-1(m) so in this case, m=1 such that θ=45°

Great explanation! Thank you!
 
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