Algebra 2 probem -- PLEASE HELP!!?
A circle is centered at the origin. The coordinates of on endpoint of a diameter are (3 + the square root of 2, -4). What are the coordinates of the other endpoint of the diameter?
Thanks so much for any light you can shed on this problem!
Comments
Center at the origin (0,0)
Diameter endpoint 1 at ( 3 + √2, -4)
The other endpoint of that diameter is easy to find; just negative the other point (as if you mirror it on both axes, really); ( - 3 - √2 , 4)
The center is the mid-point of the two end points of the diameter. So if one end-point is given, plug in the mid-point formula
Mid-point of (x,y) and (a,b)= ((x+a)/2,(y+b)/2).
In this case (0,0) is the mid-point.
So (3+sqrt 2 +a)/2=0 and (-4+b)/2=0
a and b will be your other end-points. This is the general formula.
In this special case:
You can simply use the center as a mirror and the answer is (-3-sqrt2, 4).
f(x) = x^2-6x+11 First we could make the expression x^2-6x right into a applicable sq. binomial The formula to get the consistent is (B/2)^2 so because it is (-6/2)^2 = 9 f(x) = x^2 - 6x + 9 - 9 + 11 ^-- you are able to subtract the 9 besides to maintain the balanced equation now you component the x^2 -6x + 9 and get (x-3)^2 f(x)= (x-3)^2 - 9 + 11 f(x) = (x-3)^2 + 2 it is named the "vertex sort" of a quadratic equation. interior the final sort, the sign of A will point out the path wherein the parabola opens. The vertex is (-h, ok) and the axis of symmetry is -b/2a. Voila.
(5,-7)