Question

a) A curve has equation y = (x-3)²-5 (i) Find the value of y at which the curve intersects the y-axis. (ii) Find the coordinates of the turning point of the curve. b) The sketch shows part of a graph which has equation y = ax² + bx + c. Find the values of a, b and c. ​

Answer 1

Explanation:

a)i) y = (x-3)² - 5

the curve intersects the y axis where x=0

y = (0-3)² - 5

y = 9 - 5 = 4

answer is: y = 4

a)ii) turning point (vertex) is: (3, -5)

put equation in y = ax² + bx + c form

y = (x-3)(x-3)-5 = x² - 6x + 9 - 5 = x² - 6x + 4

now find x: x = -b/2a = -(-6)/2·1 = 6/2 = 3

plug in x=3 to find y

y = (3)² - 6(3) + 4 = -5

b )sorry, can't do this one. need a 3rd point from the graph.