How do you write the quadratic function in vertex form given vertex (-4,6) and point (-1,9)?

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Answer 1 · 2017-04-24T06:45:02

#3y-18=(x+4)^2# or #(y-6)^2=3(x+4)#

A vertex form of equation is of the type

#(y-k)=a(x-h)^2# or #(x-h)=a(y-k)^2#, where #(h,k)# is the vertex.

As the vertex is #(-4,6)#, it would be either

#(y-6)=a(x+4)^2# or #(x+4)=a(y-6)^2#

Case 1

If it is #(y-6)=a(x+4)^2#, as it passes through #(-1,9)#, we have

#(9-6)=a(-1+4)^2# i.e. #9a=3# and #a=1/3# and equation is

#(y-6)=1/3(x+4)^2# i.e. #3y-18=(x+4)^2#

Case 2

If it is #(x+4)=a(y-6)^2#, as it passes through #(-1,9)#, we have

#(-1+4)=a(9-6)^2# or #3=9a# and #a=1/3# and equation is

#(x+4)=1/3(y-6)^2# or #(y-6)^2=3(x+4)#

graph{(3y-18-(x+4)^2)((y-6)^2-3(x+4))=0 [-10.96, 9.04, 0.04, 10.04]}