Given:

Expand #(x-2)^2#.

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

Simplify.

#y=2x^2-8x+8-3x#

Simplify.

#y=2x^2-11x+8# is a quadratic equation in standard form:

#y=ax^2+bx+c#,

where:

#a=2#, #b=-11#, #c=8#

**Vertex:** the maximum or minimum point of a parabola

The x-coordinate of the vertex can be calculated using the formula for the axis of symmetry:

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

#x=(-(-11))/(2*2)#

**#x=11/4# or #2.75#**

The y-coordinate is determined by substituting #11/4# for #x# in the equation and solving for #y#.

#y=2(11/4)^2-11(11/4)+8#

Simplify.

#y=2(121/16)-121/4+8#

Simplify.

#y=242/16-121/4+8#

Simplify #242/16# to #121/8#.

#y=121/8-121/4+8#

The least common denominator is #8#. Multiply #121/4# by #2/2# and multiply #8# by #8/8# in order to get equivalent fractions with #8# as the denominator. Since #n/n=1#, the numbers will change but the value will remain the same.

#y=121/8-(121/4xx2/2)+(8xx8/8)#

Simplify.

#y=121/8-242/8+64/8#

Simplify.

#y=(121-242+64)/8#

#y=-57/8# or #(-7.125)#

The vertex is #(11/4,-57/8)# or #(2.75,-7.125)#.

# graph{y=2(x^2-4x+4)-3x [-9.58, 10.42, -10.44, -0.44]} #