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How do you find a vertex in intercept form

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Quadratic functions can be written in vertex form, or


y=a\mleft(x-h\mright)^2+k

This is especially useful because the vertex of the function is found at the point (h, k).

We can find this form by completing squares, for instance, let y be:


y=x^2+bx+c

we can see that this equation is equal to


y=x^2+2((b)/(2))x+((b)/(2))^2-((b)/(2))^2+c

because


2((b)/(2))=b

and


((b)/(2))^2-((b)/(2))^2=0

However, in this form, we can see that the first 3 terms are a perfect square, that is


x^2+2((b)/(2))x+((b)/(2))^2=(x+(b)/(2))^2

hence,


\begin{gathered} y=x^2+bx+c \\ y=(x+(b)/(2))^2-((b)/(2))^2+c \end{gathered}

If we define


\begin{gathered} -((b)/(2))^2+c=k \\ \text{and} \\ h=(b)/(2) \end{gathered}

we have that


y=(x+h)^2+k

the constant a arise when you have a leading term different from 1 in x^2.

User Kjv
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