What is the vertex of # y= (x-3)^2-4x^2-x+4#?

Question

1353 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-16T20:10:01

#"Vertex"(-6/7,823/49)#

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

#"1-take derivative of the function with respect to x"#

#(d y)/(d x)=2(x-3)*1-8x-1#

#"1-equalize with zero and solve for x"#

#2(x-3)-8x-1=0#

#2x-6-8x-1=0#

#-6x-7=0#

#-6x=7#

#x=-6/7#

#"write x=-6/7 in the original equation and calculate for y"#

#y=(-6/7-3)^2-4(-6/7)^2-(-6/7)+4#

#y=(-27/7)^2-4(36/49)+6/7+4#

#y=729/49-144/49+34/7#

#y=585/49+34/7#

#y=585/49+238/49#

#y=823/49#

#y=16.8#