A1 - QUADRATIC FUNCTIONS

A1: QUADRATIC FUNCTIONS ● Finding the maximum or minimum value of a quadratic function using the method of completing the square ● Conditions for to be always positive (or always negative) 𝑦 = 𝑎 𝑥 2 + 𝑏𝑥 + 𝑐● Using quadratic functions as models 1. (a) Show that can be written as , 𝑓 ( 𝑥 ) = 3 𝑥 2 − 9 𝑥 + 7 𝑓 ( 𝑥 ) = 𝑎 ( 𝑥 − 𝑏 ) 2 + 𝑐 where and are constants to be found. 𝑎 , 𝑏 𝑐 [4] 1. (b) Hence, explain why the function is always positive. 𝑓 ( 𝑥 ) [1] 2. (a) Express in the form , where − 2 𝑥 2 − 4 𝑥 + 3 𝑎 ( 𝑥 − ℎ ) 2 + 𝑘 , and are constants. 𝑎 ℎ 𝑘 [3] 2. (b) Hence, state the maximum value of the curve . 𝑦 = − 2 𝑥 2 − 4 𝑥 + 3 [1] 3. (a) Express in the form , where and 6 𝑥 2 − 12 𝑥 + 7 𝑝 ( 𝑥 − 𝑞 ) 2 + 𝑟 𝑝 , 𝑞 𝑟 are constants to be found. [3] 3. (b) Hence find the greatest value and state the value of 6 𝑥 2 − 12 𝑥 + 7 ( ) − 1 𝑥 at which this occurs. [2] 4. (a) Express in the form where and are 𝑥 2 − 8 𝑥 + 5 𝑥 + 𝑎 ( ) 2 + 𝑏 𝑎 𝑏 constants. [2] 4. (b) Hence state the line of symmetry and the coordinates of the vertex of the curve . 𝑦 = 𝑥 2 − 8 𝑥 + 5 [2] 5. (a) Show that can be written as 𝑔 ( 𝑥 ) = − 𝑥 2 + 6 𝑥 − 14 where and are constants. 𝑔 ( 𝑥 ) = 𝑎 ( 𝑥 − 𝑏 ) 2 + 𝑐 𝑎 , 𝑏 𝑐 [3] 5. (b) Explain why the function is always negative. 𝑔 ( 𝑥 ) [2]

A1: QUADRATIC FUNCTIONS (MARKING SCHEME) 1. (a) Show that can be written as , 𝑓 ( 𝑥 ) = 3 𝑥 2 − 9 𝑥 + 7 𝑓 ( 𝑥 ) = 𝑎 ( 𝑥 − 𝑏 ) 2 + 𝑐 where and are constants to be found. 𝑎 , 𝑏 𝑐 ★ factorise so that coefficient of is 1 𝑓 ( 𝑥 ) = 3 𝑥 2 − 9 𝑥 + 7 𝑥 2 ★ you can choose to keep 7 untouched, less error 𝑓 ( 𝑥 ) = 3 𝑥 2 − 3 𝑥 ( )+ 7 𝑓 ( 𝑥 ) = 3 𝑥 2 − 2 3 2 𝑥 ( ) + 3 2 ( ) 2 − 3 2 ( ) 2 ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ + 7 𝑓 ( 𝑥 ) = 3 𝑥 − 3 2 ( ) 2 − 9 4 ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ + 7 𝑓 ( 𝑥 ) = 3 𝑥 − 3 2 ( ) 2 − 27 4 + 7 (shown) 𝑓 ( 𝑥 ) = 3 𝑥 − 3 2 ( ) 2 + 1 4 [4] 1. (b) Hence, explain why the function is always positive. 𝑓 ( 𝑥 ) ∴ Since which is more than 0. 3 𝑥 − 3 2 ( ) 2 ≥ 0 , 3 𝑥 − 3 2 ( ) 2 + 1 4 ≥ 1 4 [1] 2. (a) Express in the form , where − 2 𝑥 2 − 4 𝑥 + 3 𝑎 ( 𝑥 − ℎ ) 2 + 𝑘 , and are constants. 𝑎 ℎ 𝑘 ★ factorise so that coefficient of is 1 − 2 𝑥 2 − 4 𝑥 + 3 𝑥 2 ★ you can choose to keep 3 untouched, less error = − 2 ( 𝑥 2 + 2 𝑥 ) + 3 = − 2 𝑥 2 + 2 ( 𝑥 ) + 1 2 − 1 2 [ ]+ 3 = − 2 𝑥 + 1 ( ) 2 − 1 [ ]+ 3 = − 2 𝑥 + 2 ( ) 2 + 2 + 3 = − 2 𝑥 + 1 ( ) 2 + 5∴ − 2 𝑥 + 1 ( ) 2 + 5 [3] 2. (b) Hence, state the maximum value of the curve . 𝑦 = − 2 𝑥 2 − 4 𝑥 + 3∴ Maximum value is 5 . [1]

3. (a) Express in the form , where and 6 𝑥 2 − 12 𝑥 + 7 𝑝 ( 𝑥 − 𝑞 ) 2 + 𝑟 𝑝 , 𝑞 𝑟 are constants to be found. ★ factorise so that coefficient of is 1 6 𝑥 2 − 12 𝑥 + 7 𝑥 2 you can choose to keep 7 untouched, less error = 6 𝑥 2 − 2 𝑥 ( ) [ ]+ 7 = 6 𝑥 2 − 2 ( 𝑥 ) + 1 2 − 1 2 [ ]+ 7 = 6 𝑥 − 1 ( ) 2 − 1 [ ]+ 7 = 6 𝑥 − 1 ( ) 2 − 6 + 7 = 6 𝑥 − 1 ( ) 2 + 1∴ 6 𝑥 − 1 ( ) 2 + 1 [3] 3. (b) Hence find the greatest value and state the value of 6 𝑥 2 − 12 𝑥 + 7 ( ) − 1 𝑥 at which this occurs. Greatest value = 1 6 𝑥 2 − 12 𝑥 + 7 = 1 1 = 1 𝑥 = 1∴ 𝑥 = 1 [2] 4. (a) Express in the form where and are 𝑥 2 − 8 𝑥 + 5 𝑥 + 𝑎 ( ) 2 + 𝑏 𝑎 𝑏 constants. 𝑥 2 − 8 𝑥 + 5 = 𝑥 2 − 2 ( 4 𝑥 ) + 4 2 − 4 2 [ ]+ 5 = 𝑥 − 4 ( ) 2 − 16 + 5 = 𝑥 − 4 ( ) 2 − 11∴ 𝑥 − 4 ( ) 2 − 11 [2] 4. (b) Hence state the line of symmetry and the coordinates of the vertex of the curve . 𝑦 = 𝑥 2 − 8 𝑥 + 5 Line of symmetry, 𝑥 = 4Coordinates of the vertex of the curve, 4 , − 11( ) [2]

5. (a) Show that can be written as 𝑔 ( 𝑥 ) = − 𝑥 2 + 6 𝑥 − 14 where and are constants. 𝑔 ( 𝑥 ) = 𝑎 ( 𝑥 − 𝑏 ) 2 + 𝑐 𝑎 , 𝑏 𝑐 ★ factorise so that coefficient of is 1 𝑔 ( 𝑥 ) = − 𝑥 2 + 6 𝑥 − 14 𝑥 2 ★ you can choose to keep 14 untouched, less error 𝑔 ( 𝑥 ) = − 1 𝑥 2 − 6 𝑥 ( ) [ ]− 14 𝑔 ( 𝑥 ) = − 1 𝑥 2 − 2 ( 3 𝑥 ) + 3 2 − 3 2 [ ]− 14 𝑔 ( 𝑥 ) = − 1 𝑥 − 3 ( ) 2 − 9 [ ]− 14 𝑔 ( 𝑥 ) = − ( 𝑥 − 3 ) 2 + 9 − 14 (shown) 𝑔 ( 𝑥 ) = − ( 𝑥 − 3 ) 2 − 5 [3] 5. (b) Explain why the function is always negative. 𝑔 ( 𝑥 ) ( 𝑥 − 3 ) 2 ≥ 0 − ( 𝑥 − 3 ) 2 ≤ 0 − ( 𝑥 − 3 ) 2 − 5 ≤ − 5Hence, maximum value of , is always negative. 𝑔 ( 𝑥 ) = 5 𝑔 ( 𝑥 ) [2]