y=ax^2+bx+c

Question

If y=ax^2+bx+c passes through the points (-3,10), (0,1), and (2,15), what's the value of a+b+c?

Answer 1

y = ax² + bx + c

 

Since (-3,10) is on the graph, 9a - 3b + c = 10

Since (1,0) is on the graph,                  c =  1

Since (2,15) is on the graph,  4a + 2b + c = 15

 

Since c = 1, we then have 9a - 3b = 9

                                    4a + 2b = 14

 

Divide the first equation by 3 and the second by 2:  3a - b = 3

                                                                        2a + b = 7

 

                                               Add the equations to get 5a = 10

 

                                                                             So, a = 2

 

                                               Therefore, 3(2) - b = 3

 

                                                                       b = 3

 

a + b + c = 2 + 3 + 1 = 6

 

 

Answer 2

Hi again Anna,

 

 

If y=ax²+bx+c passes through the points (-3,10), (0,1), and (2,15), what's the value of a+b+c?

 

 

Substitute each of the points:

              y=ax²+bx+c

(-3,10)    10=a(-3)²+b(-3)+c      10=9a-3b+c

(0,1)       1=a(0)²+b(0)+c            1=c

(2,15)     15=a(2)²+b(2)+c          15=4a+2b+c

 

From the 2nd equation, we know that c=1.  so substitute the value into the 1st and 3rd equations

 

10=9a-3b+c          10=9a-3b+1        9=9a-3b

15=4a+2b+c         15=4a+2b+1       14=4a+2b

 

 

9=9a-3b       (1/3)[9=9a-3b]             3=3a-b

14=4a+2b      (1/2)[14=4a+2b]         7=2a+b

 

3=3a-b

7=2a+b

--------

10=5a

2=a

 

Substitute a=2 into 1st equation

3=3a+b

3=3(2)+b

3=6+b

-3=b

 

So far we know that:  a=2, b=-3 and c=1.

 

the value of a+b+c=2+-3+1=0

 

ANSWER: The sum of a+b+c=0.