Question

Find the equation of a parabola that has a vertex of (-2,-3) and contains the point (4,1)

Answer 1

`\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \boxed{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{-2}{ h},\stackrel{-3}{ k})\\\\ -------------------------------\\\\ \begin{cases} h=-2\\ k=-3 \end{cases}\implies y=a[x-(-2)]^2-3\implies y=a(x+2)^2-3 \\\\\\ \textit{we also know that } \begin{cases} x=4\\ y=1 \end{cases}\implies 1=a(4+2)^2-3 \\\\\\ 4=36a\implies \cfrac{4}{36}=a\implies \cfrac{1}{9}=a \\\\\\ therefore\qquad \boxed{y=\cfrac{1}{9}(x+2)^2-3}`


`\bf y=\cfrac{1}{9}(x+2)^2-3\implies y=\cfrac{1}{9}(x^2+4x+4)-3 \\\\\\ y=\cfrac{1}{9}x^2+\cfrac{4}{9}x+\cfrac{4}{9}-3\implies \stackrel{standard~form}{y=\cfrac{1}{9}x^2+\cfrac{4}{9}x-\cfrac{23}{9}}`