Rewrite f(x) = –(x + 3)2 − 1 from vertex form to standard form

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Rewrite f(x) = –(x + 3)2 − 1 from vertex form to standard form

Mathematics
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vertex form: y=a(x - h)^2+k Standard form: y = ax^2 + bx + c
so i need to plug in the numbes?

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Okay, so first thing you do is distribute the (x + 3)^2 expression. After you've distributed, multiply the expression with -1. Then, subtract -1.
im lost :/
Sorry, I think I'm explaining in a really crappy way right now. Let me reword it. So, take your vertex form expression. –1*(x + 3)^2 − 1. The first thing you want to do is simplify the term in parenthesis. Which is (x + 3)^2
F(x)=-x^2-6x-10 f(x)=x^2-6x+8 f(x)=x^2-6x-8 f(x)=x^2-8 Those are my choices btw
Okay, so first distribute. (x + 3)(x + 3).
X^2+6x+9
Plug it into the remaining parts of the equation! The equation is now –1*X^2+6x+9 − 1
Okay, so what now?
Simplify the rest of the equation. –1*X^2+6x+9 − 1 Multiply (X^2+6x+9) by -1. After you've received your answer, subtract 1 from it. That is your equation in standard form!
So it's C?
So it's C?
Nope, try again.

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