Verex form of a quadratic function
1 Answer
We try to write this fuunction $f(x)=-4x^2+4x+3$ in the vertex form $y = a(x - h)^2+ k$.
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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