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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)² + 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)² + b, the point (a,b) is the vertex of the parabola, and (a,b) are its coordinates.
y = x² + 5x
what you have to do now is complete the square.
if you have (x+p)², then square it, you get x² + 2px + p²
If the coefficient of the x²-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 x² + 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)²
If we expand this we get,
x² + 5x + 6.25
Comparing this with x² + 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,
x² + 5x = (x+2.5)² - 6.25
y = (x+2.5)² - 6.25
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.: a = -2.5, b = -6.25
Vertex = (-2.5, -6.25)
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