is y=(2x-1)^2 in vertex form/factor form?

Answer 1

A quadratic function y = f(x) is written in "vertex form" if it looks like y = a(x - h)² + k. I'd say your equation does *not* look like that, because of the coefficient on the x-term in the parentheses. To put it in vertex form, you'll need to factor out that coefficient:

y = (2x - 1)²
y = (2)²(x - 1/2)²
y = 4(x - 1/2)²

Now it's in vertex form, with a = 4, h = 1/2, and k = 0.

Hope that helps!

Answer 2

since vertex form is y = a(x - h)^2 + k

and y = (2x - 1)^2 can be written as

y = 1(2x - 1)^2 + 0

this its in vertex form, if it had been written as

y = (2x - 1)(2x - 1)

then that would be factored form.

Answer 3

'Sherman81' is completely correct except that it should be

y = 4(x-1/2)²+0

Algebraic manipulation(s) to put something in 'standard' form don't change the 'form' itself.


Doug

Answer 4

y = (2x - 1)²
y = (2x)² + 2.(2x).(- 1) + (- 1)²
y = 4x² - 4 x + 1
y = 4.(x² - x) + 1 , :4
y = (x² - x)/4 + 1/4
y = (x² - x + 1)4
y = (1/4).(x² - x + 1) or
y = x²/4 - x/4 + 1/4