SOLUTION: vertex form y=-x^2+6x+4 i need in vertex form i cant find out how to do it?

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Question 535713: vertex form y=-x^2+6x+4
i need in vertex form i cant find out how to do it?

Answer by fcabanski(1391)   (Show Source): You can put this solution on YOUR website!
y=(a(x-h)^2) + k is vertex form. The vertex is (h,k). +a is smiley, -a is frowney - that tells you if the parabola opens up or down. To convert an equation to vertex form complete the square.


y=x^2+6x+4


Step 1: move the loose number (the constant term) to the y side.


y-4=x^2 + 6x


Step 2: Factor out whatever multiplies x^2. Here it's 1, so this step does nothing except place the right side in ()'s.


y-4=(x^2 + 6x)


Step 3: Make a space after the constant on the y side, place the factor before x^2 before that space and the sign before that space is the same as the sign of the number in front of x^2. Again, since it's 1, no need for that part.


y-4+()=(x^2 + 6x)


Take 1/2 the coefficient of the x term (the number in front of x). Remember its sign. Square it. Add it to both sides inside the ().


y-4+(9)=(x^2 + 6x +9)


Multiply out the "a times the squared coefficient" part on the left-hand side (remember, in this one a = 1 so that does nothing), and convert the right-hand side to squared form. (This is where you use that sign you kept track of earlier, putting that sign in the middle of the squared expression.)


y-4+(9)=(x+3)^2


Simplify - combine like terms.


y+5=(x+3)^2


Move the constant term from the right back to the left.


y=(x+3)^2 - 5


Write in vertex form y=(a(x-h)^2) + k. In other words if the squared term is x+h write it as x-(-h). If the k term is negative, write it as + (-k).


y=(x-(-3))^2 + (-5)


Now the values for h and k are clear.


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