How do you write the quadratic function in vertex form given vertex (1,-10) and point (-3,54)?

1 Answer
Apr 23, 2017

#(y+10)=4(x-1)^2# or #(y+10)^2+1024(x-1)=0#

Explanation:

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 #(1,-10)#, it would be either

#(y+10)=a(x-1)^2# or #(x-1)=a(y+10)^2#

Case 1

If it is #(y+10)=a(x-1)^2#, as it passes through #(-3,54)#, we have

#(54+10)=a(-3-1)^2# i.e. #16a=64# and #a=4# and equation is

#(y+10)=4(x-1)^2#

Case 2

If it is #(x-1)=a(y+10)^2#, as it passes through #(-3,54)#, we have

#(-3-1)=a(54+10)^2# or #-4=4096a# and #a=-1/1024# and equation is

#(x-1)=-1/1024(y+10)^2# or #(y+10)^2+1024(x-1)=0#

graph{((y+10)-4(x-1)^2)((y+10)^2+1024(x-1))=0 [-5, 5, -80, 80]}