Junior Cert Maths Samples

_________________________________________ juniorcertsolutions.com ________________________________________ 2016 Question 14 – Higher Paper 1 Solution ( continued ) (c) (i) ( ) 2 g x ax bx c = + + To find c , find ( ) ( ) ( ) 2 0 0 6 0 180 g a c = + + = 180 c ⇒ = ° (ii) To find a and b in 2 ax bx c + + We are told the graph is symmetrical ∴ Min occurs at x = 3 since the graph of g ( x ) touches the x axis at 3 x = ∴ g ( x ) is of the form ( ) ( ) 2 2 2 3 6 9 6 9 a x a x x ax ax a − ⇒ − + ⇒ − + We know from above that 9 180 a c = = D 20, 6 120 a b a ⇒ = = = ∴ ( ) 2 20 120 180 g x x x = − + Or the longer method. The function passes through the following 3 points (0,180), (6,180) and (3,0). Graph of the function ( ) 2 g x ax bx c = + + We already know 180 c = ∴ ( ) 2 180 g x ax bx = + + ( ) 3 9 3 180 0 3 60 g a b a b = + + = ⇒ + = − ( ) 6 36 6 180 180 6 0 g a b a b = + + = ⇒ + = Solving simultaneoously, 3 60 20, 120 a a b = ⇒ = = − Comment: Difficult enough for Junior Cert.