SOLUTION: "Write each equation in vertex form (use completing the square)." Vertex Form: y=a(x-h)^2+k (h,k)=the vertex. My equation is: Y=5x^2-10x+9 By using an answer sheet supplied b

Algebra ->  Equations -> SOLUTION: "Write each equation in vertex form (use completing the square)." Vertex Form: y=a(x-h)^2+k (h,k)=the vertex. My equation is: Y=5x^2-10x+9 By using an answer sheet supplied b      Log On


   



Question 831027: "Write each equation in vertex form (use completing the square)."
Vertex Form: y=a(x-h)^2+k
(h,k)=the vertex.
My equation is:
Y=5x^2-10x+9
By using an answer sheet supplied by my teacher, I know the answer is:
Y=5(x-1)^2 +4 where the vertex is: (1,4)
However I have been unable to reach this answer I can show what I've tried below:
Y=5x^2-10x+9
-9 -9
Y-9=5x^2-10x
Y-9=5(x^2-2x)
(-2/2)^2=1---------> this step is for completing the square. The equation I used is (b/2)^2
Y-9+1=5(x^2-2x+1)
Factor
Y-8=5(x-1)^2
+8 +8
Y=5(x-1)^2+8
According to this the vertex would be: (1,8), however this is incorrect. Do you know what I'm doing wrong?
I'm sorry if this is confusing and if the formatting doesn't turn out correct, but thank you for your time.
-Val W

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
Y=5x^2-10x+9
group x's
Y=(5x^2-10x)+9
factor out 5:
Y=5(x^2-2x)+9
to complete square, we need to add 1:
(don't forget, even though we added only 1 -- EFFECTIVELY we've added 5 because of the multiplier)
Y=5(x^2-2x+1)+9 -5
Y=5(x-1)^2+4
so, vertex is at (1,4)