Question

1. Find the vertex of the parabola whose equation is y = x^2 + 6x + 2.

2. What is the line of symmetry for the parabola whose equation is y = 3x^2 + 24x - 1?

3. What is the line of symmetry for the parabola whose equation is y = x^2 + 10x + 25?

4. Find the vertex of the parabola whose equation is y = -2x^2 + 8x - 5.

5. What is the line of symmetry for the parabola whose equation is y = x^2 - 12x + 7?

6. Find the vertex of the parabola whose equation is y = x^2 - 4x + 6.

Answer 1

Answer with explanation:

We know that for any parabola equation of the type:

`y=ax^2+bx+c`

The vertex of the parabola is given by:

(h,k)

where,

`h=(-b)/(2a)\ and\ k=(4ac-b^2)/(4a)`

and the line of symmetry of the parabola is:

`x=(-b)/(2a)`

Ques 1)

We have the parabola equation as:

`y=x^2+6x+2`

i.e.

`a=1\ ,\ b=6\ and\ c=2`

Hence,

`h=(-6)/(2* 1)\\\\\\h=(-6)/(2)\\\\h=-3`

and

`k=(4* 1* 2-(6)^2)/(4* 1)\\\\\\k=(8-36)/(4)\\\\k=-7`

Hence, Vertex=(-3,-7)

and axis of symmetry is: x= -3

Ques 2)

`y=3x^2+24x-1`

We have:

a=3, b=24 and c=-1

Hence,

`h=(-24)/(2* 3)\\\\h=(-24)/(6)\\\\h=-4`

and,

`k=(4* (-1)* 3-(24)^2)/(4* 3)\\\\\\k=-49`

Hence, Vertex=( -4,-49)

and Axis of symmetry is: x= -4

Ques 3)

`y=x^2+10x+25`

`a=1\ ,b=10\ ,c=25`

Hence,

`h=(-10)/(2* 1)\\\\h=(-10)/(2)\\\\h=-5`

and,

`k=(4* (1)* 25-(10)^2)/(4* 1)\\\\\\k=0`

Hence, Vertex=( -5,0)

and Axis of symmetry is: x= -5

Ques 4)

`y=-2x^2+8x-5`

We have:

`a=-2\ ,b=8\ and\ c=-5`

Hence,

`h=(-8)/(2* (-2))\\\\h=(-8)/(-4)\\\\h=2`

and,

`k=(4* (-2)* (-5)-(8)^2)/(4* (-2))\\\\\\k=3`

Hence, Vertex=( 2,3)

and Axis of symmetry is: x= 2

Ques 5)

`y=x^2-12x+7`

We have:

`a=1\ ,b=-12\ and\ c=7`

Hence,

`h=(12)/(2* (1))\\\\h=(12)/(2)\\\\h=6`

and,

`k=(4* 1* (7)-(-12)^2)/(4* 1)\\\\\\k=-29`

Hence, Vertex=(6,-29)

and Axis of symmetry is: x= 6

Ques 6)

`y=x^2-4x+6`

We have:

`a=1\ ,b=-4\ and\ c=6`

Hence,

`h=(4)/(2* (1))\\\\h=(4)/(2)\\\\h=2`

and,

`k=(4* 1* (6)-(-4)^2)/(4* 1)\\\\\\k=2`

Hence, Vertex=(2,2)

and Axis of symmetry is: x= 2

Answer 2

1. y = x^2 + 6x + 2 = x^2 + 6x + 9 - 7 = (x + 3)^2 - 7
The vertex of the parabola whose equation is y = x^2 + 6x + 2 is (-3, -7).

2. y = 3x^2 + 24x - 1 = 3(x^2 + 8x) - 1 = 3(x^2 + 8x + 16) - 1 - 48 = 3(x + 4)^2 - 49
The line of symmetry for the parabola whose equation is y = 3x^2 + 24x - 1 is x = -4

3. y = x^2 + 10x + 25 = (x + 5)^2
The line of symmetry for the parabola whose equation is y = x^2 + 10x + 25 is x = -5

4. y = -2x^2 + 8x - 5 = -2(x^2 - 4x) - 5 = -2(x^2 - 4x + 4) - 5 + 8 = -2(x - 2)^2 + 3
The vertex of the parabola whose equation is y = -2x^2 + 8x - 5 is (2, 3).

5. y = x^2 - 12x + 7 = x^2 - 12x + 36 - 29 = (x - 6)^2 - 29
The line of symmetry for the parabola whose equation is y = x^2 - 12x + 7 is x = 6

6. y = x^2 - 4x + 6 = x^2 - 4x + 4 + 2 = (x - 2)^2 + 2
The vertex of the parabola whose equation is y = x^2 - 4x + 6 is (2, 2).