Question

Find real numbers a, b, and c so that the graph of the function y = ax² +bx+c contains the points (-1,4), (2,8), and

(0,2).

Answer 1

y= (5/3)x^2 - (1/3)x +2

Explanation:

y = ax² +bx+c

(0,2) is y-int which is c, so y = ax² + bx + 2.

(-1,4): 4 = a - b + 2

(2,8): 8 = 4a + 2b +2

4 = a - b + 2

2 = a - b (1)

8 = 4a + 2b +2

8/2 = (4a + 2b +2)/2 (divided by 2 from both sides)

4 = 2a + b + 1

3 = 2a + b (2)

(2)-(1)

3 = 2a + b

+

2 = a - b

----------------------

5 = 3a

a = 5/3

(1) 2 = a - b

2 = 5/3 - b

b= -1/3

y= (5/3)x^2 - (1/3)x +2