# Thread: finding constant a

1. ## 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...

2. ## Re: finding constant a

hi

you should post the whole question for us to help you.from here we cannot do anything.

3. ## Re: finding constant a

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

4. ## 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.

5. ## 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?

6. ## Re: finding constant a

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

7. ## Re: finding constant a

Thanks. Great forum!