Asked by erica
                what would the following equation be in vertex form. y=-9x squared+72x+81. 
I have tried for hours and i cannot figure it out
            
        I have tried for hours and i cannot figure it out
Answers
                    Answered by
            Anu
            
    Bring constant to other side: y-81=9x^2+72x
Make constant in front of x^2 term one: (y-81)/9=x^2+8x
(y/9)-9=x^2+8x
complete the square...(x+a)^2= x^2+2abx+a^2.
x^2=x^2 so
8x=2ax= so a=4
4^2=a^2=16
The equation you want to create is (x+4)^2= x^2+8x+16
previously we had (y/9)-9=x^2+8x
so add 16 to make the equation desired
(y/9)-9=x^2+8x
+16 +16
(y/9)+7=x^2+8x+16
(y/9)+7=(x+4)^2
and now solve for y:
subtract 7:(y/9)=(x+4)^2-7
multiply by 9:y=[(x+4)^2-7]9
y=9(x+4)^2-7(9)
y=9(x+4)^2-63
Vertex form: a(x-h)^2+k=y where (h,k) is the coordinate of the vertex.
You have: 9(x+4)^2-63=y where (-4,-63) is the coordinate of the vertex.
:)
    
Make constant in front of x^2 term one: (y-81)/9=x^2+8x
(y/9)-9=x^2+8x
complete the square...(x+a)^2= x^2+2abx+a^2.
x^2=x^2 so
8x=2ax= so a=4
4^2=a^2=16
The equation you want to create is (x+4)^2= x^2+8x+16
previously we had (y/9)-9=x^2+8x
so add 16 to make the equation desired
(y/9)-9=x^2+8x
+16 +16
(y/9)+7=x^2+8x+16
(y/9)+7=(x+4)^2
and now solve for y:
subtract 7:(y/9)=(x+4)^2-7
multiply by 9:y=[(x+4)^2-7]9
y=9(x+4)^2-7(9)
y=9(x+4)^2-63
Vertex form: a(x-h)^2+k=y where (h,k) is the coordinate of the vertex.
You have: 9(x+4)^2-63=y where (-4,-63) is the coordinate of the vertex.
:)
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