Question

How do you find the leading coefficients of a parabola graph?

Answer 1

well, if we use the vertex form of a parabola, we can pretty much see the parabolas all pretty much have a vertext at the origin, (0,0), now, let's take a look of a point they go by, Check the picture below.

`~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{a~is~negative}{op ens~\cap}\qquad \stackrel{a~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} h=0\\ k=0\\ \end{cases}\implies y=a(~~x-0~~)^2 + 0\hspace{4em}\textit{we also know that} \begin{cases} x=4\\ y=4 \end{cases} \\\\\\ 4=a(4-0)^2 + 0\implies 4=16a\implies \cfrac{4}{16}=a\implies \cfrac{1}{4}=a~\hfill \boxed{y=\cfrac{1}{4}x^2}`

now let's do D

`\begin{cases} h=0\\ k=0\\ \end{cases}\implies y=a(~~x-0~~)^2 + 0\hspace{4em}\textit{we also know that} \begin{cases} x=-1\\ y=-3 \end{cases} \\\\\\ -3=a(-1-0)^2+0\implies -3=a\hspace{5em}\boxed{y=-3x^2}`