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

[tex]~~~~~~\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}[/tex]

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

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

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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

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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