parabola. examples give four examples of equations of parabola all having the vertex (2,3), but one...

Parabola

kysymmetryofAxis

hkyaxEquation

aleftorarightFacingI

hxsymmetryofAxis

khxayEquation

adownoraupwFacingI

khVertex

:

)(:

:)0()0(.

:

)(:

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),(

2

2

Examples

Give four examples of equations of parabola all having the vertex (2,3), but one is facing upward, one downward, one to the right and one to the left. Identify the axis of symmetry.

Solution

Vertex (2,3)

2:

3)2(5:

:.

2:

3)2(5:

:.

.

2

2

xsymmetryofAxis

xyEquation

downFacingb

xsymmetryofAxis

xyEquation

upFacinga

I

3:

2)3(5:

:.

3:

2)3(5:

:.

.

2

2

ysymmetryofAxis

yxEquation

leftFacinga

ysymmetryofAxis

yxEquation

rightFacinga

II

Examples (1) – (4)

Identify the parabola having the given equation and its axis of symmetry and graph it.

Decide Which parabola have the following Equations & Graph:

542.4

152.3

2)1(3.2

2.1

2

2

2

2

yyx

xxy

yx

yx

Decide Which Parabola have the Given Equations & Graph:

)0,0(

)(0:

0)0(2

:Re

2.1

2

2

isvertxThe

axisxtheylinetheissymmetryofaxistheThe

rightthetofacingparabolaaisThis

yx

formndardtastheinequationthewrite

yx

Finding points of the graphx=2y2

y x (x,y)

-2 8 (8,-2)

-1 2 (2,-1)

0 0 (0,0)

1 2 (2,1)

2 8 (8,2)

3 18 (18,3)

Decide Which Parabola have the Given Equations & Graph:

)1,2(:

1:

2)1(3.2 2

isvertexThw

ylinetheissymmetryofaxistheThe

leftthetofacingparabolaaisThis

yx

Finding points of the graphx=-3(y-1)2+2

y x (x,y)

-2 -25 (-25,-2)

-1 -10 (-10,-1)

0 -1 (-1,0)

1 2 (2,1)

2 1 (-1,2)

3 -10 (-10,3)

What are the x-intercepts and the y-intercepts?

1835.03

211658.1

3

21,

)3

21,0()

3

21,0(

3

21

3

21

3

2)1(

2)1(32)1(30

:0,

1232)10(3

:0,

2)1(3.2

2

22

2

2

andWhere

andataxisythectsserteingraphThe

yyy

yy

xletweterceptsinythefindTo

x

yletweterceptsinxthefindTo

yx

Decide Which Parabola have the Given Equations & Graph:

)16,1(

1:

16)1(

151)1(

15]1)1[(

15)2(

:

152.3

2

2

2

2

2

isvertexThe

xissymmetryofaxisThe

downwardfacingparabolaaofequationanisThis

x

x

x

xxy

squarethecompletingbythatdoWe

formndardstatheinequationthewriteretoneedWe

xxy

Finding points of the graphy=-x2-2x+15 = - (x+1) 2 + 16

x y (x,y)

-2 15 (-2,15)

-1 16 (-1,16)

0 15 (0,15)

1 12 (1,12)

2 7 (2,7)

3 0 (3,0)

-5 0 (-5,00

-3 12 (-3,12)

What are the x-intercepts and the y-intercepts?

)0,5()0,3(

54134141

16)1(

16)1(0:,

)15,0(

1516)10(:,

16)1(

2

2

2

2

andataxisxthecrossgraphThe

xorxx

x

xgetwezerobeyLetting

ataxisythecrossgraphThe

ygetwezerobexLetting

xy

Decide Which Parabola have the Given Equations & Graph:

)1,3(:

1:

3)1(2

52)1(2

5]1)1[(2

5)2(2

542

:

542.4

2

2

2

2

2

2

vertexThe

yissymmetryofaxisThe

rightthetofacingparabolaaofequationanisThis

y

y

y

y

yyx

squarethecompletingbythatdoWe

formndardstathenequationithewriteretoneedWe

yyx

Finding points of the graphx=-2y2+4x+5 = 2 (y+1) 2 + 3

y x (x,y)

-4 21 (21,-4)

-3 11 (11,-3)

-2 5 (5,-2)

-1 3 (3,-1)

0 5 (5,0)

1 11 (11,1)

2 21 (21,2)

What are the x-intercepts and the y-intercepts?

axisyhecrossnotdoesgraphThe

solutionrealnohaswhichy

yequationthegetwexLetting

aaxisxthecrossesgraphThe

xgetweyLetting

yx

,3)1(2

3)1(20:,0

)0,5(

53)10(2:,0

3)1(2

2

2

2

2

More Examples

Example (1)Graph y=x2 - 6x + 7

The graph is facing upward. Why?

y= x2 - 6 x + 7 = (x- 3 )2 – 9 + 7= (x- 3 )2 – 2

The vertex of the parabola is (3 , - 2 )

The intersection with the y-axis is(0,7)y=x2 - 6x + 7 Why?

The intersections with the x-axis: (3+√2 , 0 ) and (3-√2 , 0 ) Why?

7.552.50-2.5

7.5

5

2.5

0

-2.5

-5

x

y

x

y

)2,3(

23 23

7

Example (2)Graph y=2x2 + 2x + 3

The graph is concave upward. Why?y=2[x2 +x] + 3 = 2[(x+½)2–¼] + 3= 2(x+½)2–½ + 3 = 2(x+½)2 + 5/2 The vertex of the parabola is ( - ½ , 5/2 )

The intersection with the y-axis is (0,3) Why?

No intersections with the x-axis: Why?

1.250-1.25-2.5-3.75

8

6

4

2

0

x

y

x

y

3),( 25

21

Example (3)Graph y=-x2 - 4x + 12

The graph is concave downward. Why?y= -(x2 +4x ) + 12=-[(x+ 2 )2 – 4] + 12= -(x+2)2 + 16 The vertex of the parabola is (-2 , 16 )

The intersection with the y-axis is (0,12) Why?

y=-x2 - 4x + 12=-(x2 +4x - 12)=-(x+6)(x-2)The intersections with the x-axis: (2 , 0 ) and (-6 , 0 ) Why?

52.50-2.5-5-7.5

15

10

5

0

-5

-10

x

y

x

y

)16,2(

12

6 2

QuestionDo it now!

Do not see the next slide before you do it!

Graph y=x2 + 2x - 3

Solution y=x2 +2x - 3

The graph is concave upward. Why?y=x2 +2x – 3 = = (x+1)2 – 1 – 3 = (x+1)2 - 4 The vertex of the parabola is (-1 , -4 )

The intersection with the y-axis is (0,-3)y = x2 + 2x - 3=(x+3)(x-1) why?

The intersections with the x-axis: (-3 , 0 ) and (1 , 0 ) Why?

52.50-2.5-5

30

25

20

15

10

5

0

-5x

y

x

y

3)4,1( 31