so, we know the parabola runs through [tex]\bf (\stackrel{x}{0}~,~\stackrel{y}{3}),(\stackrel{x}{1}~,~\stackrel{y}{4}),(\stackrel{x}{-1}~,~\stackrel{y}{-6})[/tex]
now, since we know is a parabola, we could use the form of say y = ax² + bx + c, to find the integers a, b and c, with those points,
[tex]\bf (\stackrel{x}{0}~,~\stackrel{y}{3}),(\stackrel{x}{1}~,~\stackrel{y}{4}),(\stackrel{x}{-1}~,~\stackrel{y}{-6})\qquad \qquad \qquad y=ax^2+bx+c\\\\
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\begin{cases}
\stackrel{y}{3}=a\stackrel{x^2}{0^2}+b\stackrel{x}{0}+c\\
\stackrel{y}{4}=a\stackrel{x^2}{1^2}+b\stackrel{x}{1}+c\\
\stackrel{y}{-6}=a\stackrel{x^2}{(-1)^2}+b\stackrel{x}{(-1)}+c
\end{cases}\implies
\begin{cases}
3=\boxed{c}\\\\
4=a+b+c\\\\
-6=a-b+c
\end{cases}[/tex]
[tex]\bf \stackrel{second~equation~substitution}{4=a+b+\boxed{3}}\implies 1=a+b\implies 1-b=\underline{a}
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\stackrel{third~equation~substitution}{-6=(\underline{1-b})-b+\boxed{3}}\implies -10=-2b
\\\\\\
\cfrac{-10}{-2}=b\implies 5=b\\\\
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\textit{since we know c = 3 and b = 5, then }\stackrel{second~equation}{4=a+5+3}\implies -4=a\\\\
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y=-4x^2~~+5x~~+3[/tex]