Question

Write the equation of a parabola, in standard form, that goes through these points:

(0, 3) (1, 4) (-1, -6)

Answer 1

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`