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

1 Answer
May 16, 2017

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

Explanation:

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