SOLUTION: I am not sure on how to ask about this but on my paper for homework it says Rewrite in vertex form. and the 1st problem is y=x^2+5x i dont understand on how to do this co

Question 30118: I am not sure on how to ask about this but on my paper for homework it says Rewrite in vertex form. and the 1st problem is y=x^2+5x
i dont understand on how to do this could you please help me??
Ashley
Answer by Fermat(136) (Show Source): You can put this solution on YOUR website!
y=x^2+5x
This eqn (any quadratic eqn) can be written in the form,
y = (x-a)� + b
Quadratic eqns are parabolas, and you can think of the "sharp end" as the vertex.
When you write a quadratic/parabola in the form: y = (x-a)� + b, the point (a,b) is the vertex of the parabola, and (a,b) are its coordinates.
y = x� + 5x
what you have to do now is complete the square.
if you have (x+p)�, then square it, you get x� + 2px + p�
If the coefficient of the x�-term is 1, then the coefficient of the x-term is twice the constant value, p, in the (x+p) bracket.
So if you have x� + 5x, and the coefficent of the x-term is 5, then the coefficient of the p-term is half this, so p must be 2.5.
Your bracketed-term must then be: (x+2.5)�
If we expand this we get,
x� + 5x + 6.25
Comparing this with x� + 5x, we can see that we have 6.25 too much. So we have to subtract that.
Here is all the action in one fell swoop,
x� + 5x = (x+2.5)� - 6.25
y = (x+2.5)� - 6.25
===================
.: a = -2.5, b = -6.25
Vertex = (-2.5, -6.25)
======================
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