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

Given:

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

First you need to get the equation into standard form.

Expand #(x+3)^2# using the FOIL method.
https://www.mathsisfun.com/definitions/foil-method.html

#y=4x^2-2x+x^2+6x+9#

Collect like terms.

#y=(4x^2+x^2)+(-2x+6x)+9#

Combine like terms.

#y=5x^2+4x+9# is a quadratic equation in standard form:

#ax^2+bx+c#,

where:

#a=5#, #b=4#, #c=9#

The vertex is the maximum or minimum point of a parabola. Since #a>0#, the vertex is the minimum point of this parabola, and the parabola opens upward.

The x-coordinate of the vertex is the same as the axis of symmetry for a quadratic equation in standard form. The formula is:

#x=(-b)/(2a)#

#x=(-4)/(2*5)#

#x=-4/10#

Simplify.

#x=-2/5# or #-0.4#

To calculate the y-coordinate of the vertex, substitute #-2/5# for #x# in the equation and solve for #y#.

#y=5(-2/5)^2+4(-2/5)+9#

#y=5(4/25)-8/5+9#

#y=20/25-8/5+9#

Simplify #20/25# to #4/5#.

#y=4/5-8/5+9#

Multiply #9# by #5/5# to get an equivalent fraction with #5# as the denominator.

#y=4/5-8/5+9xx5/5#

#y=4/5-8/5+45/5#

Simplify.

#y=41/5# or #8.2#

The vertex is #(-2/5,41/5)# or #(-0.4,8.2)#.

graph{y=5x^2+4x+9 [-11.72, 13.59, 5.72, 18.38]}