Mathematics, 07.01.2020 18:31 haileywilliams1

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.

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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).

Answer from: karlaperez7309

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,

$h=\dfrac{-b}{2a}\ and\ k=\dfrac{4ac-b^2}{4a}$

and the line of symmetry of the parabola is:

$x=\dfrac{-b}{2a}$

Ques 1)

We have the parabola equation as:

y=x^2+6x+2

i.e.

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

Hence,

$h=\dfrac{-6}{2\times 1}\\\\\\h=\dfrac{-6}{2}\\\\h=-3$

and

$k=\dfrac{4\times 1\times 2-(6)^2}{4\times 1}\\\\\\k=\dfrac{8-36}{4}\\\\k=-7$

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,

$h=\dfrac{-24}{2\times 3}\\\\h=\dfrac{-24}{6}\\\\h=-4$

and,

$k=\dfrac{4\times (-1)\times 3-(24)^2}{4\times 3}\\\\\\k=-49$

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,

$h=\dfrac{-10}{2\times 1}\\\\h=\dfrac{-10}{2}\\\\h=-5$

and,

$k=\dfrac{4\times (1)\times 25-(10)^2}{4\times 1}\\\\\\k=0$

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,

$h=\dfrac{-8}{2\times (-2)}\\\\h=\dfrac{-8}{-4}\\\\h=2$

and,

$k=\dfrac{4\times (-2)\times (-5)-(8)^2}{4\times (-2)}\\\\\\k=3$

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,

$h=\dfrac{12}{2\times (1)}\\\\h=\dfrac{12}{2}\\\\h=6$

and,

$k=\dfrac{4\times 1\times (7)-(-12)^2}{4\times 1}\\\\\\k=-29$

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,

$h=\dfrac{4}{2\times (1)}\\\\h=\dfrac{4}{2}\\\\h=2$

and,

$k=\dfrac{4\times 1\times (6)-(-4)^2}{4\times 1}\\\\\\k=2$

Hence, Vertex=(2,2)

and Axis of symmetry is: x= 2

Answer from: Quest

answer: a.

step-by-step explanation: the answer is a. because, you need to converge the fractions to find a solution.

Answer from: Quest

d. corresponding side lengths in similar triangles are proportional in length.

step-by-step explanation:

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