PLS PLS help! Type an equation of a parabola that is moved down 2 units, stretched vertically by a factor of 3, and moved left 5 units.

9514 1404 393

  y = 3x^2 +30x +69

Step-by-step explanation:

Transformations work this way:

  g(x) = k·f(x) . . . . vertical stretch by a factor of k

  g(x) = f(x -h) +k . . . . translation (right, up) by (h, k)

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So, the translation down 2 units will make the function be ...

  f(x) = x^2  ⇒  f1(x) = f(x) -2 = x^2 -2

The vertical stretch by a factor of 3 will make the function be ...

  f1(x) = x^2 -2  ⇒  3·f1(x) = f2(x) = 3(x^2 -2)

The horizontal translation left 5 units will make the function be ...

  f2(x) = 3(x^2 -2)  ⇒  f2(x +5) = f3(x) = 3((x +5)^2 -2)

The transformed function equation can be written ...

  y = 3((x +5)^2 -2) = 3(x^2 +10x +25 -2)

  y = 3x^2 +30x +69

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The attachment shows the original function and the various transformations. Note that the final function is translated down 6 units from the original. That is because the down translation came before the vertical scaling.

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.

find the vertical and horizontal asymptotes of the function f x = 5 x − 1 . to find the vertical asymptote, equate the denominator to zero and solve for . since the degree of the polynomial in the numerator is less than that of the denominator, the horizontal asymptote is .

step-by-step explanation: