x
5
4
3
2
1
Section 4.3
1) f(x) = (x – 3) 2 + 4 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 xcolumn 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

-3
-4
-5 x
-1
-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
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 xcolumn when I find points. In this case I will put (-3) in the middle of my x-column.
-4
-2
4 h(x) or y
4
-2
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

-4
-5
-6 x
-2
-3
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 xcolumn 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
1
(−2 + 4)
(−3 + 4)
(−4 + 4)
2
2
2
− 6
− 6
− 6
2
1
2
1
(−5 + 4) 2 − 6
− 6
2
(−6 + 4) 2
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)
-4
-3 x
-7
-6
-5 f(x)
-3
-2.25
-2
-2.25
-3 computation
1
− (−7 + 5) 2
4
1
− (−6 + 5) 2
4
1
− (−5 + 5) 2
4
1
− (−4 + 5) 2
4
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.
-4
-5 x
-1
-2
-3
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 xcolumn 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 f(x) = (x+3) 2 – 4
Answer to part a 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 k(x) = (x-2) 2 – 2
Answer to part a 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 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) - 3 – 8
Factor and simplify f(x) = 2(x+2) 2 – 11
Answer to part a 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 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) + 4 - (-9)
Factor and simplify f(x) = -(x-3) 2 +13
Answer to part a 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 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) 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