( ) (3, (−2)2 through the point = 5) . Substituting x, y 5 = + 1 a these values for and we can solve for x, y, h, k, a: = + 5 4a 1 (3 5 = − 1)2 + (−3) a = 4 4a (2)2 = 1 5 = − 3 a a The quadratic function is = − 5 4a 3 ( ) ( = − 2)2 + 1 = − 4 + 5 . f x x x2 x = 8 4a = 2 a have = 1 and = −3 . The also passes h k The quadratic function is ( ) ( = 2 − 1)2 − 3 = 2 − 4 − 1 . f x x x2 x
5 = a (−2)2 + 1 5 = 4a + 1 4 = 4a 1 = a The quadratic function is f ( x ) = ( x − 2)2 + 1 = x2 − 4 x + 5 . have h = 1 and k = −3 . The also passes through the point ( x, y ) = (3, 5) . Substituting these values for x, y, h, and k, we can solve for a: 5 = a (3 − 1)2 + (−3) 5 = a (2)2 − 3 5 = 4a − 3 8 = 4a 2 = a The quadratic function is f ( x ) = 2 ( x − 1)2 − 3 = 2 x2 − 4 x − 1 .
Downloadable-Solution-Manual-for-Precalculus-Concepts-Through-1.docx