# finding constant a

• Sep 12th 2011, 11:36 AM
Franciscus
finding constant a
Can you give me hint...Where to start ?

I want to find constant a for function

((ax^2 - 6x + 4)/(x^2 - x - 2))

when lim x to 2...
• Sep 12th 2011, 11:46 AM
anonimnystefy
Re: finding constant a
hi

you should post the whole question for us to help you.from here we cannot do anything.
• Sep 12th 2011, 11:56 AM
CaptainBlack
Re: finding constant a
Quote:

Originally Posted by Franciscus
Can you give me hint...Where to start ?

I want to find constant a for function

((ax^2 - 6x + 4)/(x^2 - x - 2))

when lim x to 2...

What have you done?

I presume you want the limit as x approaches 2 to be finite?

If so you need (x-2) to be a factor of the numerator (it is already a factor of the denominator).

CB
• Sep 12th 2011, 12:03 PM
Franciscus
Re: finding constant a
Find a value for constant a in a function f(x) = ((ax^2 - 6x + 4) divided (x^2 - x - 2)) so that the limit exists, when the x approaches 2.
• Sep 12th 2011, 01:10 PM
Franciscus
Re: finding constant a
Thanks for quick answer CB. Yes I meant finite. I'm from Finland, so I'm not familiar with all the notions.

So in denominator I got (x-2)(x+1). Can I use these values for x in numerator or do I have to factor it other way to find out a?
• Sep 12th 2011, 07:15 PM
CaptainBlack
Re: finding constant a
Quote:

Originally Posted by Franciscus
Thanks for quick answer CB. Yes I meant finite. I'm from Finland, so I'm not familiar with all the notions.

So in denominator I got (x-2)(x+1). Can I use these values for x in numerator or do I have to factor it other way to find out a?

As I said you need \$\displaystyle x-2\$ to be a factor of the numerator (to cancel the \$\displaystyle x-2\$ factor in the denominator)

For \$\displaystyle x-2\$ to be a factor of the numerator you need \$\displaystyle a(2)^2-6(2)+4=0\$

CB
• Sep 13th 2011, 08:26 AM
Franciscus
Re: finding constant a
Thanks. Great forum!