Question

Select the quadratic function that has a vertex at (3, –5).

Question 3 options:

y=(x−3)2−5

y=(x−3)2+5

y=(x+3)2−5

y=(x−5)2+3

y=(x−5)2−3

y=(x+5)2+3

Answer 1

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

we know the vertex is at h = 3 and k = -5

$\stackrel{vertex}{(\stackrel{h}{3}~~,~~\stackrel{k}{-5})}\qquad \implies y=(x-3)^2-5$