Question

Please help I need to show my work

Answer 1

from the picture above, we can see that the focus point of the parabola is at (2 , -4) and its vertex at (2 , -5), so the "p" distance from each other is just 1 unit, since the vertical parabola is opening upwards, that means "p" is positive or just p=1.

`\textit{vertical parabola vertex form with focus point distance} \\\\ 4p(y- k)=(x- h)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h,k+p)}\qquad \stackrel{directrix}{y=k-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{p~is~negative}{op ens~\cap}\qquad \stackrel{p~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill`

`\begin{cases} h=2\\ k=-5\\ p=1 \end{cases}\implies 4(1)(~~y-(-5)~~) = (~~x-2~~)^2 \\\\\\ 4(y+5)=x^2+4x+4\implies 4y+20=x^2+4x+4 \\\\\\ 4y=x^2+4x-16\implies {\Large \begin{array}{llll} y=\cfrac{1}{4}x^2+x-4 \end{array}}`