Hi,can anyone help me out with this. I would like to check my child homework to see if he completed correctly.

Question

1. Find the constant term that should be added to make the following expression a perfect-square trinomial. x^2+2x

2. Solve by completing the square.
2x^2-6x-3=0

3. Find the x-intercepts.
y=x^2-4x+4

4. Identify the axis of symmetry, create a table of values and include the axis of symmetry.
y= -x^2+3x-3

Thanks to anyone that can help!

Answer 1

1. Find the constant term that should be added to make the following expression a perfect-square trinomial. x²+2x
x² + 2x + c²
(x + c)(x + c)
(x + 1)(x + 1)
(x+1)²
x² + 2x + 1
c = 1

2. Solve by completing the square.
2x²-6x-3=0
(2x² - 6x) - 3
2(x² - 3x) - 3
--->-3*1/2 = (-3/2)² = 9/4
2(x² - 3x + 9/4) - 3 - (9/4*2)
2(x² - 3x + 9/4) - 3 - 9/2
2(x² - 3x + 9/4) - 15/2
0 = 2(x² - (3/2))² - 15/2
"standard form" from completing the square.

3. Find the x-intercepts.
y=x²-4x+4
y = 0, It'll cross the x-axis
0 = x² - 4x + 4
0 = (x - 2)(x - 2)
0 = x - 2
x = 2
So x-intercepts are at: (2, 0) only.

4. Identify the axis of symmetry, create a table of values and include the axis of symmetry.
y= -x²+3x-3

x l y
-------
-2 l -13
-1 l -7
0 l -3 <--y-intercept
1 l -1
***axis of symmetry**** x = 1.5
2 l -1
3 l -3
4 l -7
5 l - 13

This is an down facing parabola. ∩
axis of symmetry is x = ? Because this is a vertical line.
x = -b/2a ---> if your equation is ax² + bx + c = y
x = -3/2(-1)
x = 3/2 or 1.5 "axis of symmetry"

Answer 2

1. Find the constant term that should be added to make the following expression a perfect-square trinomial. x^2+2x


Add ((1/2)(b))^2.

which is ((1/2) * 2) ^2 - > 1

x^2+2x +1

+1 is the constant term,

2. Solve by completing the square.

2x^2-6x-3=0

2x^2 -6x = 3

2(x^2 - 3x) = 3
add ((1/2)(b))^2.

(1/2 * 3 ) ^2 -> (3/2)^ 2 = 9/4

2(x^2 - 3x+ 9/4) = 3 + (9/4 * 2)
2(x - 3/2)^2 = 3 + 9/4
2(x - 3/2)^2 = 12/4 + 9/4
2(x - 3/2)^2 - 21/4 = 0


3. Find the x-intercepts.
y=x^2-4x+4

(x+2)(x-2)

x = -2 and x = 2 are your intercepts

4. Is a parabola.

-x(x-3) - 3 = 0

x = h = axis-of-symmetry = 3

Answer 3

Question 1
x² + 2x
= (x² + 2x + 1) - 1
= (x + 1)² - 1

Question 2
x² - 3x = 3 / 2
x² - (3 / 2)x + 9 / 16 = 3 / 2 + 9 / 16
(x - 3 / 4)² = 33 / 16
(x - 3/4) = ±√33 / 4
x = 3 / 4 ± √33 / 4
x = (1/4) (3 ±√33)

Question 3
y = x² - 4x + 4
y = (x - 2)²
Intercepts at (0 , 4) and (2 , 0)

Qustion 4
y = - (x² - 3x + 3)
y = - (x² - 3x + 9 / 4 - 9 / 4 + 3)
y = - [ (x - 3/2)² + 3/4 ]
Axis of symmetry is x = 3/2
x**** 3/2*****0***** 1***** 2 etc
y*** -3/4****- 3*** - 1****- 1

Answer 4

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