Using the completing-the-square method, rewrite f(x) = x2 + 4x − 1 in vertex form.
f(x) = (x + 2)2 + 1
f(x) = (x + 2)2
f(x) = (x + 2)2 + 4
f(x) = (x + 2)2 − 5

- anonymous

- katieb

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- anonymous

welcome to open study

- anonymous

tanks

- anonymous

yeah

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## More answers

- anonymous

can you help me with this problem

- anonymous

what does it mean by "rewrite f(x) = x2 + 4x − 1 in vertex form."

- anonymous

- anonymous

- marigirl

factorizing,
so basically when you expand it out again, you should have your original equation.

- anonymous

but how do i turn fx to vetex form

- anonymous

vertex

- anonymous

o i didnt see where you said factor sorry

- marigirl

vertex form also means turning point on graph. i.e at which point on the x axis will the parabola have its minimum point.
Best way to go about is start factorizing the equation given to you f(x) = x^2 + 4x − 1
May start factorising first part - x^2+4x .. what will that get us to

- anonymous

x(x+4) ???

- marigirl

let try it another way .. i am going to factorise it to
\[f(x)= (x+2)^{2}\]
This will ensure I have it in vertex form... now i need to manipulate it a little so that my final equation will look like \[f(x)=x ^{2}+4x-1\]
so now, what constant value will i need to add to the top equation to make it same as the second equation ..

- anonymous

i dont know 2?

- marigirl

see what u get when you expand f(x)=(x+2^2:
we get x^2+4x+4 ..... but i need x^2+4x-1 .. so what number do i need to add/minus so i wil get to x^2+4x-

- anonymous

1

- marigirl

i need to minus 5

- anonymous

where u get that from lol im so confused

- marigirl

(x+2)^2-5

- marigirl

sorry i dont think i helped...

- anonymous

oooo ok i understand wher i am i was confused with the vertex form i mixed it with a different formula

- anonymous

no actually you did i just thought to hard

- marigirl

ok, well you can always expand the answers given to you and see which is the original equation

- anonymous

ok thanksss

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