how do you turn this (vertex form) to standard form?
y= -(x-1)^2+5

Question

how do you turn this (vertex form) to standard form?
y= -(x-1)^2+5

anonymous · Answer

@FutureMathProfessor

anonymous · Answer

Follow through with (x-1)^2 and write out the full expression

anonymous · Answer

would it be -(x-1)(x-1) or how does the negative work?

anonymous · Answer

@amistre64 please help

amistre64 · Answer

the negative will effect the product, changes signs is all

amistre64 · Answer

-(4)^2 = -16

amistre64 · Answer

-(x-1)^2

-(x^2-2x+1)

-x^2 +2x -1

amistre64 · Answer

then add the 5 in :)

anonymous · Answer

so -x^2-2x+1 to start with? then +5 to the +1 ?

anonymous · Answer

oops -1*

amistre64 · Answer

-(x^2-2x+1)  , swap out ALL the signs

amistre64 · Answer

your distributing a -1 thru the paranthsis

anonymous · Answer

do you distribute a negative 1 even to the second set of parenthesis?

amistre64 · Answer

youve already taken care of the second set ... by "foiling" or however you want to describe it

amistre64 · Answer

$$-(x-1)^2$$
$$-(x-1)(x-1)$$
$$-(x^2-2x+1)$$
$$-x^2--2x-+1$$
$$-x^2+2x-1$$

anonymous · Answer

basically I am trying to find the zero's of the equation

amistre64 · Answer

finding the zeros does not need to be determined from the standard form

do you have to convert this to standard form, or just find the zeroes?

anonymous · Answer

just find the zero's

amistre64 · Answer

then let y=0
$$0= -(x-1)^2+5$$
$$(x-1)^2=5$$
$$x-1=\pm\sqrt{5}$$
$$x=1\pm\sqrt{5}$$

anonymous · Answer

thanks you helped out a lot and answered all my questions! :)

amistre64 · Answer

:)  good luck