factorise 1-16(1-y)^2
please help me
difference of squares
= (1 + 4(1-y) )(1- 4(1-y))
= (1 + 4 - 4y)(1-4 + 4y)
= (5-4y)(4y - 3)
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)
given a GP below, find the values of m, and the next 2 terms:
m, m+3, m^2+3
Help me please
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)
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
To factorize the expression 1 - 16(1 - y)^2, we can start by applying the formula a^2 - b^2 = (a + b)(a - b). In this case, let's observe that a = 1 and b = 4(1 - y). Hence, we can rewrite the expression as follows:
1 - 16(1 - y)^2 = 1 - [4(1 - y)]^2
Now, let's apply the formula a^2 - b^2 = (a + b)(a - b):
= (1 + 4(1 - y))(1 - 4(1 - y))
= (1 + 4 - 4y)(1 - 4 + 4y)
= (5 - 4y)(-3 + 4y)
Hence, the factorized form of the expression 1 - 16(1 - y)^2 is (5 - 4y)(-3 + 4y).