Paul George M.

asked • 04/20/18

T(x)=ax^2+bx+c,T(0)=-4,T(1)=-2,T(2)=6

How do I find a, b and c?

1 Expert Answer

By:

Mark M. answered • 04/20/18

Tutor
4.9 (952)

Math Tutor--High School/College levels
About this tutor ›
About this tutor ›
T(x) = ax2+bx+c
 
Since T(0) = -4, c = -4
 
So, T(x) = ax2 + bx - 4
 
Since T(1) = -2, a + b - 4 = -2
 
Since T(2) = 6, 4a + 2b - 4 = 6
 
We have the system of equations:  a + b = 2
                                                  4a + 2b = 10
 
Divide the second equation by 2 to get a + b = 2
                                                       2a + b = 5
 
    Subtracting the equations, -a = -3.  So, a = 3
 
Since a = 3 and a + b = 2, b = -1.
 
T(x) = 3x2 - x - 4
 
Upvote • 0 Downvote
More
Report

Paul George M.

Thank you very much for your answer. I've got the same result. 
Report

04/20/18

Paul George M.

I did like that:
when x=0 than c=-4 y intercept 
when x=1 than  a+b-4=-2 and from this statement I found a=2-b
and when x=2 this equation   T(x) = ax2+bx+c  looks like 4(2-b)+2b-4=6. From here I found b=-1
and I substituted -1 for b in this eq. to find the real a value a+b-4=-2  a=3
then T(x)=3x^2-x-4
Report

04/20/18

Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.
Most questions answered within 4 hours.

OR

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.