Which equation matches the quadratic equation displayed in the graph?

a) y = x2 - 1
b) y = x2 + 1
c) y = (x + 1)2
d) y = (x - 1)2

Step-by-step explanation:

vertex: (m,n) = (0,1)

f(x) = a(x-m)² + n = a(x-0)²+1

y = x² +1

Tip: you could pick any point on the curve and plug the x value in the given equations to see  if you get the y value

eg. if you pick (x,y) = (-2,5) as your point, the right equation would give you 0 which is y.

x²+ 1 =y; -2²+1 = 4+1=5; so when x is -2, y is 5

the correct answer is a) they believe they have a "just cause."

islamic terrorists justify their violent actions because they believe they have a "just cause."

every act of terrorism of islamic sects is done “in the name of allah”. that justifies their actions. islam people who commit those terrorist attacks are not afraid to die. that is why the attacks have horrible consequences. they know they are going to die and they are proud to do it because it is in the name of allah, so it is justified and is for a “good cause.”

since the very beginning of the conflict with israel, islam is fighting a “sacred war”, respecting the words stated in the qur’an. they fight against the domination of the western countries and to defend their occupied territories.

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.