Section 2.2 Check Point Exercises ( ) 3. 4 f x x = −x2 + +1 Step 1: The parabola opens down because 0. a < Step 2: find the vertex: ( ) ( 1. f x x = − 4 b 4 −1)2 + 2 x =− =− = a =−1 2a 2(−1) 2 h=1 k =4       ( )  f x 4(2) 5 = + +1 = The vertex is (2, 5). − (2) f = −22  1  4 x − + Step 1: The parabola opens down because 0. a < Step 2: find the vertex: (1, 4) Step 3: find the x-intercepts: Step 3: find the x-intercepts: 0 4x = −x2 + +1 ( 0 4 x = − −1)2 + b −b ± 2 − 4ac x = 2a ( x −1)2 = 4 −4 42 4(−1)(1) − x −1 = ± 2(−1) ±2 1 x x = ± = 2 = 3 or x x = −1 Step 4: find the y-intercept: 20 x = −4 ± −2 (0) (0 f = − −1)2 + 4 3 2 5 x = = ± The are and . x-intercepts x x ≈ −0.2 ≈ −4.2 Step 5: The axis of symmetry is = 1. x (0) 4(0) f = −02 + +1 = Step 4: find the y-intercept: 1 Step 5: The axis of symmetry is = 2. x ( ) ( 2. f x x = − 2)2 +1 Step 1: The parabola opens up because 0. 4. ( 4 a f x) x2 x > = −16 +1000 Step 2: find the vertex: (2, 1) a. = 4. The parabola opens upward and has a a Step 3: find the x-intercepts: minimum value. ( 0 2)2 x = − +1 ( 2)2 x − = 16 −1 −b b. 2 x = = = 2 x − = 2a 8 −1
Section 2.2 Check Point Exercises 3. f ( x ) = −x2 + 4 x +1 Step 1: The parabola opens down because a < 0. Step 2: find the vertex: 1. f ( x ) = − ( x −1)2 + 4 b 4 x =− =− = 2 a =−1 h=1 2 k =4 2a 2(−1)      f ( x ) = −  x − 1  + 4   f (2) = −22 + 4(2) +1 = 5 The vertex is (2, 5). Step 1: The parabola opens down because a < 0. Step 2: find the vertex: (1, 4) Step 3: find the x-intercepts: 0 = − ( x −1)2 + 4 Step 3: find the x-intercepts: 0 = −x2 + 4x +1 −b ± b 2 − 4ac x = 2a ( x −1)2 = 4 x −1 = ±2 x = 1 ± 2 −4 ± x = 42 − 4(−1)(1) 2(−1) x = 3 or x = −1 Step 4: find the y-intercept: x = −4 ± 20 −2 f (0) = − (0 −1)2 + 4 = 3 x = 2 ± 5 The x-intercepts are x ≈ −0.2 and x ≈ −4.2 . Step 5: The axis of symmetry is x = 1. Step 4: find the y-intercept: f (0) = −02 + 4(0) +1 = 1 Step 5: The axis of symmetry is x = 2. 2. f ( x ) = ( x − 2)2 +1 Step 1: The parabola opens up because a > 0. Step 2: find the vertex: (2, 1) Step 3: find the x-intercepts: 0 = ( x − 2)2 +1 4. f ( x) = 4 x2 −16 x +1000 a. a = 4. The parabola opens upward and has a minimum value. ( x − 2)2 = −1 −b 16 b. x = = = 2 x − 2 = −1 2a 8
Precalculus-Essentials-5th-Edition-Blitzer-Solutions-Manual.docx