Section a) f(x-3)+4 = (x 3) the (-3) in the parenthesis moves right 3, the +4 moves up 4

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1 Section 4.3 1a) f(x-3)+4 = (x 3) the (-3) in the parenthesis moves right 3, the +4 moves up 4 Answer 1a: f(x-3)+4 = (x 3) The graph has the same shape as f(x) = x 2, except it is shifted right 3 units and up 4 units. 1b) I will change the sign of the number in the parenthesis and put that number in the middle of my x- column when I find points. In this case I will put 3 in the middle of my x-column. x f(x) or y Computation, use calculator to get y - column 5 8 (5-3) (4-3) (3-3) (2-3) (1-3) c) Domain (, ) Range [4, ) (see me for help if you need some finding the domain and range) 1d) The graph is increasing (3, ) and decreasing from (, 3) This is the part of the graph that is increasing This is the part of the graph that is decreasing

2 1e) The graph does not have a high point so it has no local maximum 1f) The low point is the local minimum. We say there is a local minimum at x = 3 and the local minimum value is y = 4 3a) 2f(x+3) 4 = 2(x+3) 2-4 The (+3) in the parenthesis moves left 3, the -4 moves down 4, the 2 in front of the parenthesis stretches the graph vertically (makes skinnier). Answer #3a: 2f(x+3) 4 = 2(x+3) 2 4 The graph has the same shape as f(x) = x 2, except it is shifted left 3 units and down 4 units and it is narrower (stretched). 3b) I will change the sign of the number in the parenthesis and put that number in the middle of my x- column when I find points. In this case I will put (-3) in the middle of my x-column. x h(x) or y Computation, use calculator to get y - column (-1+3) (-2+3) (-3+3) (-4+3) (-5+3) 2-4 3c) Domain (, ) Range [ 4, ) (see me for help if you need some finding the domain and range)

3 3d) The graph is increasing ( 3, ) and decreasing from (, 3) This is the part of the graph that is increasing This is the part of the graph that is decreasing 3e) The graph does not have a high point so it has no local maximum 3f) The low point is the local minimum. We say there is a local minimum at x = -3 and the local minimum value is y =- 4 5a) 1 2 f(x + 4) 6 = 1 2 (x + 4)2 6 the (+4) in the parenthesis moves left 4, the -6 moves down 6, the 1/2 in front of the parenthesis compresses the graph vertically (makes wider). Answer #5a: 1 2 f(x + 4) 6 = 1 2 (x + 4)2 6 The graph has the same shape as f(x) = x 2, except it is shifted left 4 units and down 6 units and the graph is wider (compressed) than f(x) = x 2.

4 5b) I will change the sign of the number in the parenthesis and put that number in the middle of my x- column when I find points. In this case I will put (-4) in the middle of my x-column. x h(x) or y Computation, use calculator to get y - column ( 2 + 4) ( 3 + 4) ( 4 + 4) ( 5 + 4) ( 6 + 4)2 6 5c) Domain (, ) Range [ 6, ) (see me for help if you need some finding the domain and range) 5d) The graph is increasing ( 4, ) and decreasing from (, 4) This is the part of the graph that is increasing This is the part of the graph that is decreasing 5e) The graph does not have a high point so it has no local maximum

5 5f) The low point is the local minimum. We say there is a local minimum at x = -4 and the local minimum value is y =- 6 7a) -2f(x) + 3 = -2x you may think of this as -2(x-0) if it helps the +3 moves the graph up 3 units, the (-) in front of the 2 reflects the graph over the x-axis and the 2 stretches the graph vertically. The 2 part of the -2 makes the graph narrow. Answer #7a: -2f(x) + 3 = -2x The graph has the same shape as f(x) = x 2 except it is moved up 3 and reflected over the x-axis, and it is narrower (or stretched) 7b) x m(x) or y computation (2) (1) (0) (-1) (-2) c) Domain (, ) Range (, 3] (see me for help if you need some finding the domain and range) 7d) The graph is increasing from (, 0) and decreasing from (0, ) This is where the graph is increasing This is where the graph is decreasing

6 7e) The local maximum occurs at the vertex. We say there is a local maximum at x = 0 and the local maximum value is y = 3. 7f) There is no local minimum or local minimum value. 9) 1 4 f(x + 5) 2 = 1 4 (x + 5)2 2 9a) the (-) in front of the ¼ reflects the graph over the x-axis. The +5 in the parenthesis shifts the graph 5 units to the left. The (-2) moves the graph down 2. The ¼ compresses the graph or you may say it makes it wider. Answer #9a: 1 4 f(x + 5) 2 = 1 4 (x + 5)2 2 The graph is the same as g(x) = x 2, except moved left 5, down 2 and reflected over the x-axis. The graph is wider, or compressed

7 9b) x f(x) computation ( 7 + 5) ( 6 + 5) ( 5 + 5) ( 4 + 5) ( 3 + 5)2 2 9c) Domain (, ) Range (, 2] (see me for help if you need some finding the domain and range) 9d) The graph is increasing from (, 5) and decreasing from ( 5, ) Region where graph is increasing Region where graph is decreasing

8 9e) The local maximum occurs at the vertex. We say there is a local maximum at x = -5 and the local maximum value is y = -2. 9f) There is no local minimum or local minimum value. 11a) 2f(x+3)+2 = 2(x +3) a) the (+3) in the parenthesis moves left 3, the +4 moves up 4, the 2 in front of the parenthesis stretches the graph vertically (makes skinnier Answer #11a: 2f(x+3)+2 = 2(x +3) 2 +4 The graph has the same shape as g(x) = x 2, except it is shifted left 3 units and up 4 units and is narrower (stretched). 11b) I will change the sign of the number in the parenthesis and put that number in the middle of my x- column when I find points. In this case I will put (-3) in the middle of my x-column. x h(x) or y Computation, use calculator to get y - column (-1+3) (-2+3) (-3+3) (-4+3) (-5+3) c) Domain (, ) Range [4, ) (see me for help if you need some finding the domain and range)

9 11d) The graph is increasing ( 3, ) and decreasing from (, 3) Region where graph is increasing Region where graph is decreasing 11e) The graph does not have a high point so it has no local maximum 11f) The low point is the local minimum. We say there is a local minimum at x = -3 and the local minimum value is y = 4 13a) f(x) = x 2 + 6x + 5 Rewrite group the x s f(x) = (x 2 + 6x ) + 5 find C = ( 1 2 6)2 =9 Add and subtract C f(x) = (x 2 + 6x + 9) Factor and simplify f(x) = (x+3) 2 4 Answer to part a f(x) = (x+3) b)

10 15a) k(x) = x 2 4x + 2 Rewrite group the x s k(x) = (x 2-4x ) + 2 find C = ( 1 2 4)2 =4 Add and subtract C k(x) = (x 2-4x +4) Factor and simplify k(x) = (x-2) 2 2 Answer to part a k(x) = (x-2) b) 17a) f(x) = 2x 2 +8x 3 Rewrite group the x s f(x) = (2x 2 +8x ) - 3 Factor out GCF of 2 f(x) = 2(x 2 + 4x ) - 3 find C = ( 1 2 4)2 = 4 Add and subtract C inside parenthesis and 2C outside f(x) = 2(x 2 + 4x +4) Factor and simplify f(x) = 2(x+2) 2 11 Answer to part a f(x) = 2(x+2) b)

11 19a) f(x) = -x 2 + 6x + 4 Rewrite group the x s f(x) = (-x 2 + 6x ) + 4 Factor out GCF of -1 f(x) = -(x 2 6x ) + 4 find C = ( 1 2 6)2 = 9 Add and subtract C inside parenthesis and -C outside f(x) = -(x 2 6x + 9) (-9) Factor and simplify f(x) = -(x-3) Answer to part a f(x) = -(x-3) b) 21) k(x) = -2x x - 7 Rewrite group the x s k(x) = (-2x x ) - 7 Factor out GCF of -2 k(x) = -2(x 2 6x ) - 7 find C = ( 1 2 6)2 = 9 Add C inside parenthesis and subtract -2C outside k(x) = -2(x 2 6x + 9) 7 (-18) Factor and simplify k(x) = -2(x-3) Answer to part a k(x) = -2(x-3) 2 +11

12 21b) 23) f(x) = -3x 2 12x + 1 Rewrite group the x s f(x) = (-3x 2 12x ) + 1 factor out 3 f(x) = -3(x 2 + 4x ) + 1 Find C = ( 1 2 4)2 =4 Add 4 inside and subtract -3(4) or -12 outside the parenthesis f(x) = -3(x 2 + 4x + 4) +1 (-12) factor and simplify answer to part a: f(x) = -3(x+2) b)

13 25) use the formula f(x) = a(x-h) 2 + k Use the vertex (1, -4) as the values for h, k that is make h = 1 and k = -4 f(x) = a(x- 1) 2 4 f(x) = a(x-1) 2-4 now use the other point (-2,14) plug in -2 for x and replace the f(x) with 14 and solve for a. 14 = a(-2-1) = a(-3) = 9a =9 a 2 = a Write answer: f(x) = 2(x-1) ) use the formula f(x) = a(x-h) 2 + k Use the vertex (-1, 5) as the values for h, k that is make h = -1 and k = 5 f(x) = a(x- -1) f(x) = a(x+1) now use the other point (0,2) plug in 0 for x and replace the f(x) with 2 and solve for a. 2 = a(0+1) = a(1) = a = a Write answer: f(x) = -3(x+1) 2 +5

14 29) use the formula f(x) = a(x-h) 2 + k Use the vertex (-2, 6) as the values for h, k that is make h = -2 and k = 6 f(x) = a(x- -2) f(x) = a(x+2) now use the other point (2,14) plug in 2 for x and replace the f(x) with 14 and solve for a. 14 = a(2+2) = 16a = 16a 8/16 = a ½ = a Write answer: f(x) = 1 (x + 2 2) ) use the formula f(x) = a(x-h) 2 + k Use the vertex (-2, 3) as the values for h, k that is make h = -2 and k = 3 f(x) = a(x- -2) f(x) = a(x+2) now use the other point (2,-1) plug in 2 for x and replace the f(x) with -1 and solve for a. -1 = a(2+2) = 16a = 16a -4/16 = a -1/4 = a Write answer: f(x) = 1 (x + 4 2)2 + 3

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