# Verex form of a quadratic function

## 1 Answer

We will do this by factoring the coefficient of $x^2$ first and then completing the square:

\[f(x)=-4x^2+4x+3=-4(x^2-x-\frac{3}{4})\]

\[=-4(x^2-x+\frac{1}{4}-\frac{3}{4}-\frac{1}{4})\]

\[=-4(x^2-x+\frac{1}{4}-1)\]

\[=-4((x-\frac{1}{2})^2-1)\]

\[=-4(x-\frac{1}{2})^2+4,\]

which is in the vertex form.

Poincare

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