Question

Please help! Is my working right or have I made a mistake?

Answer 1

let's do it hmmm this way

what is the equation of a line whose vertex is at (-1 , -9) and has a solution or x-intercept at x = 2, namely it passes through (2 , 0)?

`~~~~~~\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=-1\\ k=-9\\ \end{cases}\implies y=a(~~x-(-1)~~)^2 + (-9)\hspace{4em}\textit{we also know that} \begin{cases} x=2\\ y=0 \end{cases} \\\\ 0=a(~~2-(-1)~~)^2 + (-9)\implies 0=a(3)^2-9\implies \cfrac{9}{3^2}=a\implies 1=a \\\\[-0.35em] ~\dotfill\\\\ y=1( ~~ x-(-1) ~~ )^2+(-9)\implies y=(x+1)^2-9\implies \boxed{y=x^2+2x-8}`

now, is that unique? well, nope, because we could equally have a horizontal parabola with the same vertex and the same x-intercept.

`~~~~~~\textit{horizontal parabola vertex form} \\\\ x=a(y- k)^2+ h\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{a~is~negative}{op ens~\supset}\qquad \stackrel{a~is~positive}{op ens~\subset} \end{cases} \\\\[-0.35em] ~\dotfill`

`\begin{cases} h=-1\\ k=-9\\ \end{cases}\implies x=a(~~y-(-9)~~)^2 + (-1)\hspace{4em}\textit{we also know that} \begin{cases} x=2\\ y=0 \end{cases} \\\\ 2=a(~~0-(-9)~~)^2 + (-1)\implies 3=81a\implies \cfrac{3}{81}=a\implies \cfrac{1}{27}=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill~ x=\cfrac{1}{27}(y+9)^2 -1~\hfill~`

now, to the extent that the form is expected to be y = ax² + bx + c, the solution is unique to that form.

notice the picture below of both parabolas.