math

factorise 1-16(1-y)^2
please help me

asked by Robert
  1. difference of squares

    = (1 + 4(1-y) )(1- 4(1-y))
    = (1 + 4 - 4y)(1-4 + 4y)
    = (5-4y)(4y - 3)

    posted by Reiny
  2. I am very good at maths, I am estimated to get an A+, but I hate turning point questions.

    How do you find the turning point when given the quadratic function?

    f(x)=(x+3)^2 - 16
    Write down the co-ordinates of the turning point on the grap of y=f(x)

    posted by Bobby
  3. given a GP below, find the values of m, and the next 2 terms:
    m, m+3, m^2+3
    Help me please

    posted by Robert
  4. For a quadratic function , the turning point would be the vertex of the parabola.
    since your equation is written in vertex form
    f(x) = (x+3)^2 - 16
    the vertex is (-3,-16) , which would be your turning point.

    If your quadratic is written in the form
    f(x) = ax^2 + bx + c
    there is a quick way to find the vertex
    the x of the vertex is -b/(2a)
    once you have that x, sub it into the equation to get the y
    e.g.
    y = 3x^2 + 9x - 11
    the x of the vertex is -9/6 = -3/2
    then y = 3(9/4) + 9(-3/2) - 11
    = 24/4 - 54/4 - 44/4
    = -37/2
    the vertex or turning point is (-3/2 , -37/2)

    posted by Reiny
  5. if m , m+3, m^2 + 3 are in geometric progression, then

    (m+3)/m = (m^2 + 3)/(m+3)
    m^3 + 3m = m^2 + 6m + 9
    m^3 - m^2 - 3m - 9 = 0

    normally cubics are hard to solve, but I tried factors of 9
    and found m = 3 to work

    so the numbers are 3, 6, 12
    so each number is double its previous one, r = 2
    the next 2 numbers are 24 , 48

    posted by Reiny

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