# math

factorise 1-16(1-y)^2

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

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.

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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