Question

Select the quadratic function graphed below.


y=(x+2)2−4

y=2(x+2)2−4

y=4(x+2)2−4

y=(x−2)2−4

y=2(x−2)2−4

y=4(x−2)2−4

Answer 1

$~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases}$

from the graph we can see that the vertex is at h = -2 and k = - 4, we also know that the parabola runs through the point (-1 , 0)

$\stackrel{vertex}{(\stackrel{h}{-2}~~,~~\stackrel{k}{-4})}\qquad y=a[x-(-2)]^2-4\implies \underline{y=a(x+2)^2-4} \\\\\\ \begin{cases} x=-1\\ y=0 \end{cases}\qquad \implies 0=a(-1+2)^2-4\implies 0=a-4\implies 4=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \underline{y=4(x+2)^2-4}~\hfill$