How do you write Y=-6(x-2)^2-9Y=−6(x−2)2−9 in standard form?

1 Answer
M7chael
Oct 6, 2017

y=-6x^2+24x-33y=−6x2+24x−33

Explanation:

Standard form for a quadratic is y=ax^2+bx+cy=ax2+bx+c

The best way to do this is to simplify using order of operations.

First exponents,

(x-2)^2=(x-2)(x-2)(x−2)2=(x−2)(x−2)
Using FOIL x*x=x^2, x*-2=-2x, -2*x=-2x, and -2*-2=4x⋅x=x2,x⋅−2=−2x,−2⋅x=−2x,and−2⋅−2=4

So (x-2)(x-2)=x^2-2x-2x+4=x^2-4x+4(x−2)(x−2)=x2−2x−2x+4=x2−4x+4

Next multiplication,

-6(x^2-4x+4)=-6x^2+24x-24−6(x2−4x+4)=−6x2+24x−24

Finally subtraction,

-6x^2+24x-24-9=-6x^2+24x-33−6x2+24x−24−9=−6x2+24x−33

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