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Sagot :
Quadratic functions can be written in vertex form, or
[tex]y=a\mleft(x-h\mright)^2+k[/tex]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:
[tex]y=x^2+bx+c[/tex]we can see that this equation is equal to
[tex]y=x^2+2(\frac{b}{2})x+(\frac{b}{2})^2-(\frac{b}{2})^2+c[/tex]because
[tex]2(\frac{b}{2})=b[/tex]and
[tex](\frac{b}{2})^2-(\frac{b}{2})^2=0[/tex]However, in this form, we can see that the first 3 terms are a perfect square, that is
[tex]x^2+2(\frac{b}{2})x+(\frac{b}{2})^2=(x+\frac{b}{2})^2[/tex]hence,
[tex]\begin{gathered} y=x^2+bx+c \\ y=(x+\frac{b}{2})^2-(\frac{b}{2})^2+c \end{gathered}[/tex]If we define
[tex]\begin{gathered} -(\frac{b}{2})^2+c=k \\ \text{and} \\ h=\frac{b}{2} \end{gathered}[/tex]we have that
[tex]y=(x+h)^2+k[/tex]the constant a arise when you have a leading term different from 1 in x^2.
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