For x²+bx+c, to complete the square, you add (half of b)²
b=8 in this case, so you add 4², which is 16
x²+8x+16-16-4
y=(x+4)²-20 is the final form.
Of the four x-coordinates to choose only 1/√(11) belongs can belong to the unit circle.
The other three x-coordinates are greater than 1, then they are out of the unit circle.
The unit circle formula is x^2 +y^2 = 1
Then to find the y-coordinate given the x-coordinate you can solve for y from that formula:
y^2 = 1 - x^2
y = (+/-)√(1-x^2)
Substitute the value of x
y = (+/-)√{1 - [1/√(11)]^2} = (+/-) √{(1 - 1/11} =(+/-) √ {(11 -1)/11 =(+/-)√(10/11) ≈ +/- 0.95
|4x + 4| = 15
First, let's get rid of the absolute value signs to get the first solution.
4x + 4 = 15
Subtract 4 from both sides
4x = 11
Divide each side by 4
x = 11/4 or 2.75
Now, let's get the solution by changing 15 to -15.
4x + 4 = -15
Subtract 4 from both sides
4x = -19
Divide both sides by 4
x = -19/4 or -4.75
