A parabola opening up or down has vertex

 Write its equation in vertex form.

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vertex (0,-3)

P (-4,-5)

\(normal\ form\\ y=ax^2-3\\ -5=a\cdot16-3\\ a=-\dfrac{2}{16}\\ a=-\dfrac{1}{8}\\ \color{blue}f(x)=-\dfrac{1}{8}x^2-3\)

\( a=-\frac{1}{8}\\ d=0\\ e=-3\\ vertex\ form\\ f(x)=a(x−d)^2+e\\ \color{blue}f(x)=-\frac{1}{8}(x-0)^2-3\)

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