+0  
 
0
62
1
avatar

If f(0) = 0, f(1) = 2, f(2) = 9, and ax^2+bx+c=0, what is the value of f(3)?

 Aug 13, 2021
 #1
avatar+9312 
+1

Assuming  f(x) = ax2 + bx + c,

 

f(0)  =  a(02) + b(0) + c  =  c     and so   c  =  0

f(1)  =  a(12) + b(1) + c  =  a + b + c  =  a + b + 0  =  a + b     and so   a + b  =  2

f(2)  =  a(22) + b(2) + c  =  4a + 2b + c  =  4a + 2b + 0  =  4a + 2b     and so     4a + 2b  =  9

 

Now since  a + b = 2  ,  we can subtract  b  from both sides to get   a = 2 - b

 

Now we can substitute  2 - b  in for  a  into the equation  4a + 2b = 9  to get:

 

4(2 - b) + 2b  =  9          distribute  4  to both terms in parenthesees

8 - 4b + 2b  =  9          combine like terms

8 - 2b  =  9          subtract  8  from both sides of the equation

-2b  =  1          divide both sides of the equation by  -2

b  =  -1/2

 

Now we can substitute  -1/2  in for  b  into the equation  a + b = 2  to get:

 

a + b  =  2

a - 1/2  =  2

a  =  2 + 1/2

a  =  5/2

 

And so  f(x)  =  \(\frac52\)x2  - \(\frac12\)x  +  0

 

Now to find  f(3) ,  plug in  3  for  x  into the function.

 

f(3)  =  \(\frac52\)(3)2  - \(\frac12\)(3)  +  0   =   \(\frac{45}2\)  -  \(\frac32\)   =   \(\frac{42}{2}\)   =   21

 Aug 13, 2021

41 Online Users

avatar