advertisement

Section 4.3 1) f(x) = (x – 3) 2 + 4 x 5 4 3 2 1 a) the (-3) in the parenthesis moves right 3, the +4 moves up 4 The graph has the same shape as g(x) = x 2 , except it is shifted right 3 units and up 4 units. b) 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. f(x) or y 8 5 4 5 8 Computation, use calculator to get y - column (5-3) 2 + 4 (4-3) 2 + 4 (3-3) 2 + 4 (2-3) 2 + 4 (1-3) 2 + 4 1c) 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

x -1 -2 -3 -4 -5 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 3) h(x) = 2(x +3) 2 -4 3a) 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). You are not responsible for stating the effect the 2 has on the graph. The graph has the same shape as g(x) = x 2 , except it is shifted left 3 units and down 4 units. 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. h(x) or y 4 -2 -4 -2 4 Computation, use calculator to get y - column 2(-1+3) 2 - 4 2(-2+3) 2 - 4 2(-3+3) 2 - 4 2(-4+3) 2 - 4 2(-5+3) 2 - 4 3c) Domain (−∞, ∞) Range [−4, ∞) (see me for help if you need some finding the domain and range) 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

x -2 -3 -4 -5 -6 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 5) 𝑔(𝑥) = 1 2 (𝑥 + 4) 2 − 6 5a) 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). You are not responsible for stating the effect the 1/2 has on the graph. The graph has the same shape as f(x) = x 2 , except it is shifted left 4 units and down 6 units. 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. h(x) or y -4 -5.5 -6 -5.5 -4 Computation, use calculator to get y - column 1 2 1 (−2 + 4) (−3 + 4) 2 2 − 6 − 6 2 1 2 1 2 1 2 (−4 + 4) (−5 + 4) (−6 + 4) 2 2 2 − 6 − 6 − 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 x 2 1 0 -1 -2 5e) The graph does not have a high point so it has no local maximum 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 7) m(x) = -2x 2 + 3 you may think of this as -2(x-0) 2 + 3 if it helps 7a) 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. You are not responsible for stating the effect the 2 has on the graph. The graph has the same shape as f(x) = x 2 except it is moved up 3 and reflected over the x-axis 7b) m(x) or y -5 1 3 1 -5 computation -2(2) 2 + 3 -2(1) 2 + 3 -2(0) 2 + 3 -2(-1) 2 + 3 -2(-2) 2 + 3

7c) 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 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) 𝑓(𝑥) = − 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 vertically. You don’t need to describe the effect of the ¼. The graph is the same as g(x) = x 2 , except moved left 5, down 2 and reflected over the x-axis.

9b) x -7 -6 -5 -4 -3 f(x) -3 -2.25 -2 -2.25 -3 computation − 1 4 (−7 + 5) 2 − 1 4 (−6 + 5) 2 − 1 4 (−5 + 5) 2 − 1 4 (−4 + 5) 2 − 1 4 (−3 + 5) 2 − 2 − 2 − 2 − 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

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. x -1 -2 -3 -4 -5 9f) There is no local minimum or local minimum value. 11) b(x) = 2(x +3) 2 +4 11a) 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). You are not responsible for stating the effect the 2 has on the graph. The graph has the same shape as g(x) = x 2 , except it is shifted left 3 units and up 4 units. 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. h(x) or y 12 6 4 6 -12 Computation, use calculator to get y - column 2(-1+3) 2 +4 2(-2+3) 2 +4 2(-3+3) 2 + 4 2(-4+3) 2 + 4 2(-5+3) 2 + 4 11c) Domain (−∞, ∞) Range [4, ∞) (see me for help if you need some finding the domain and range)

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) + 5 – 9 Factor and simplify Answer to part a f(x) = (x+3) 2 – 4 f(x) = (x+3) 2 – 4 13b)

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

19a) f(x) = -x 2 + 6x + 4 Rewrite – group the x’s f(x) = (-x 2 + 6x ) + 4 Factor out GCF of -1 find C = ( 1 2 ∗ −6) 2 = 9 f(x) = -(x 2 – 6x ) + 4 Add and subtract C inside parenthesis and -C outside Factor and simplify Answer to part a f(x) = -(x 2 – 6x + 9) + 4 - (-9) f(x) = -(x-3) 2 +13 f(x) = -(x-3) 2 +13 19b) 21) k(x) = -2x 2 + 12x - 7 Rewrite – group the x’s k(x) = (-2x 2 + 12x ) - 7 Factor out GCF of -2 find C = ( 1 2 ∗ −6) 2 = 9 k(x) = -2(x 2 – 6x ) - 7 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) 2 +11 Answer to part a k(x) = -2(x-3) 2 +11

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) 2 + 13 23b)

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) 2 -4 14 = a(-3) 2 – 4 14 = 9a -4 18 =9 a 2 = a Write answer: f(x) = 2(x-1) 2 - 4 27) 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) 2 + 5 f(x) = a(x+1) 2 + 5 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) 2 + 5 2 = a(1) 2 + 5 2 = a + 5 -3 = a Write answer: f(x) = -3(x+1) 2 +5

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) 2 + 6 f(x) = a(x+2) 2 + 6 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) 2 + 6 14 = 16a + 6 8 = 16a 8/16 = a ½ = a Write answer: f(x) = 1 2 (𝑥 + 2) 2 + 6 31) 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) 2 + 3 f(x) = a(x+2) 2 + 3 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) 2 + 3 -1 = 16a + 3 -4 = 16a -4/16 = a -1/4 = a Write answer: f(x) = − 1 4 (𝑥 + 2) 2 + 3