Question

Find the equation of the form y=ax²+bx+c whose graph passes through the points (1,6), (3, 20), and (−2,15).

Answer 1

Answer:

$c = \dfrac{25}{4}$

$b = \dfrac{-13}{8}$

$a = \dfrac{11}{8}$

Step-by-step explanation:

given ,

equation y=ax²+bx+c

passing through points (1,6), (3, 20), and (−2,15).

then these points will satisfy the equation

at (1,6)

y  = a x²+b x+c

6 = a(1)² + b (1) + c

a + b + c = 6------(1)

at (3 , 20)

y  = a x²+b x+c

20 = a(3)² + b (3) + c

9 a + 3 b + c = 20------(2)

at (−2,15)

y  = a x²+b x+c

15 = a(-2)² + b (-2) + c

4 a -2 b + c = 15------(3)

solving equation (1),(2) and (3)

a = 6 - b - c

9 (6 - b - c)+ 3 b + c = 20

6 b + 7 c = 34-------(4)

4 (6 - b - c) -2 b + c = 15

2 b + c = 3----------(5)

on solving equation (4) and (5)

$c = \dfrac{25}{4}$

$b = \dfrac{-13}{8}$

$a = \dfrac{11}{8}$