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## #1 2009-08-25 23:51:00

Greaterpathmagician
Member
Registered: 2009-06-16
Posts: 32

### Changing the values of a, b and c

Hello

How can you determine the affects of changing the a, b and c values in quadratic, cubic, exponential and recprical equations without graphing them.

I know the basics that in quadratics changing the 'a' stretches sthe graph vertically, changing the 'b' value translates the graph horizontally and changin the graph in 'c' translates the graph vertically.

In my assighment I had to describe the effects of the exponential function y= 3^x which changes to y= 4^x

The answers is: The graph rises faster in quad 1, falls faster in quad 2. y intercept remains 1

I know how the y-interecepts remains one but how do you determine if the graph rises faster in quad one ect.

The other one was y= (x+2)^2-3 was altered to y= (x-3)^2 +2

The answer: The turning point moves from (-2, -3) to (3,2). Graph moves 5 units right and 5 units up.

How do you determine those decriptions.

How would you describe the affects for other functions such as cubics and reciprical functions.

Thank you

Thank you for the help for my previous questions, this is the best forum to get help for maths.

Last edited by Greaterpathmagician (2009-08-25 23:53:01)

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## #2 2009-08-26 00:13:17

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: Changing the values of a, b and c

Hi Greaterpathmagician;

In my assighment I had to describe the effects of the exponential function y= 3^x which changes to y= 4^x

The answers is: The graph rises faster in quad 1, falls faster in quad 2. y intercept remains 1

I know how the y-interecepts remains one but how do you determine if the graph rises faster in quad one ect.

I am assuming that since you aren't allowed to use graphing, I will have to explain without using any kind of higher math. The simplest way is to just think like this.

In quadrant 1: x is positive
In quadrant 2: x is negative.

Since 4 is greater than 3  4^x is always greater than 3^x in Quadrant 1:

4^(1/2) >3^(1/2)
4^1 > 3^1
4^2 > 3^2
4^3 > 3^3
etc.

The function is always increasing or as you say rising faster.

Now when the graph is in quadrant 2 the  x's are negative. Negative powers go in the denominator like this:
4^(-2) = 1/(4^2)
4^(-3) = 1/(4^3)

the graph takes these values:

4^(-1/2) < 3^(1/2)
4^(-1)    < 3^(-1)
4^(-2)    < 3^(-2)
4^(-3)    < 3^(-3)
etc.

In quadrant 2:  4^(-x) < 3^(x)
So the graph is always decreasing or as you say falling faster.

Is any of that clear?

Last edited by bobbym (2009-08-26 00:50:10)

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #3 2009-08-26 00:30:48

juriguen
Member
Registered: 2009-07-05
Posts: 59

### Re: Changing the values of a, b and c

Hi!

I would say that, if for example you have:

you can write

So, changing c and b effectively moves the origin of coordinates for the function

The same happens with

And if you check the derivative:

you see that the parameter a controls the rate of change of y.

Then, for the exponentiation function, you can check the derivative as well

In this case, for 0 < a < 1, increasing a makes the function to decrease slower (lna < 0, and increasing a makes the lna to be smaller in module. But be careful, because in this case the function itself is decreasing). And for 1 < a < inf, increasing a makes the function to increase faster (lna > 0, and increasing a makes lna to grow).

Hope it helps!
Jose

Last edited by juriguen (2009-08-26 00:37:18)

Make everything as simple as possible, but not simpler. -- Albert Einstein

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