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

Answer:[tex]c = \dfrac{25}{4}[/tex][tex]b = \dfrac{-13}{8}[/tex][tex]a = \dfrac{11}{8}[/tex]Step-by-step explanation:given , equation y=ax²+bx+cpassing through points (1,6), (3, 20), and (−2,15).then these points will satisfy the equationat (1,6)y  = a x²+b x+c6 = a(1)² + b (1) + ca + b + c = 6------(1)at (3 , 20) y  = a x²+b x+c20 = a(3)² + b (3) + c9 a + 3 b + c = 20------(2)at (−2,15) y  = a x²+b x+c15 = a(-2)² + b (-2) + c4 a -2 b + c = 15------(3)solving equation (1),(2) and (3)a = 6 - b - c9 (6 - b - c)+ 3 b + c = 206 b + 7 c = 34-------(4)4 (6 - b - c) -2 b + c = 152 b + c = 3----------(5)on solving equation (4) and (5)[tex]c = \dfrac{25}{4}[/tex][tex]b = \dfrac{-13}{8}[/tex][tex]a = \dfrac{11}{8}[/tex]