advertisement

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