Mathematics: Write the equation of a parabola, in standard form, that goes through these points: (0, 3) (1, 4) (-1, -6)

Write the equation of a parabola, in standard form,

Ответ:

cjjjjjjjjjjjjj

15.03.2022

So, we know the parabola runs through $\bf (\stackrel{x}{0}~,~\stackrel{y}{3}),(\stackrel{x}{1}~,~\stackrel{y}{4}),(\stackrel{x}{-1}~,~\stackrel{y}{-6})$

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,

$\bf (\stackrel{x}{0}~,~\stackrel{y}{3}),(\stackrel{x}{1}~,~\stackrel{y}{4}),(\stackrel{x}{-1}~,~\stackrel{y}{-6})\qquad \qquad \qquad y=ax^2+bx+c\\\\
-------------------------------\\\\
\begin{cases}
\stackrel{y}{3}=a\stackrel{x^2}{0^2}+b\stackrel{x}{0}+c\\
\stackrel{y}{4}=a\stackrel{x^2}{1^2}+b\stackrel{x}{1}+c\\
\stackrel{y}{-6}=a\stackrel{x^2}{(-1)^2}+b\stackrel{x}{(-1)}+c
\end{cases}\implies 
\begin{cases}
3=\boxed{c}\\\\
4=a+b+c\\\\
-6=a-b+c
\end{cases}$

$\bf \stackrel{second~equation~substitution}{4=a+b+\boxed{3}}\implies 1=a+b\implies 1-b=\underline{a}
\\\\\\
\stackrel{third~equation~substitution}{-6=(\underline{1-b})-b+\boxed{3}}\implies -10=-2b
\\\\\\
\cfrac{-10}{-2}=b\implies 5=b\\\\
-------------------------------\\\\
\textit{since we know c = 3 and b = 5, then }\stackrel{second~equation}{4=a+5+3}\implies -4=a\\\\
-------------------------------\\\\
y=-4x^2~~+5x~~+3$

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Write the equation of a parabola, in standard form, that goes through these points: (0, 3) (1, 4) (-1, -6)

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