How do you graph #y= 3( x - 2) ^ { 2} + 1#?

[Make equation easier to read]

y = #3(x - 2)^2# + 1
= #3(x^2 - (2)(x)(2) + 2^2)# + 1
= #3(x^2 - 4x + 4)# + 1
= #3x^2# - 12x + 12 + 1
= #3x^2# - 12x + 13

[find the highest/lowest point of the equation by using #dy/dx#]

dy/dx = 6x - 12
the gradient (known as #dy/dx#) of a top/bottom of a curve line is 0

when #dy/dx# = 0
#dy/dx# = 0 (easy algebra job :) )
x = 12/6 = 2

the point x coordinate is 2
now we have to identify the point's y coordinate:
y = #3x^2# - 12x + 13
= #3(2)^2# - 12(2) + 13
= 3(4) - 24 + 13
= 1

so the coordinate of this lowest/highest point is (2,1)

[identify then trend of the parabola by using #dy/dx#]
when x = 0 (0<2)
y = 6x - 12 = 6(0) - 12 = -6
-6 is negative, which means the line is going down (if you read from left to right) when x<2

when x = 4 (4>2)
y = 6x - 12 = 6(4) - 12 = 24-12=12
12 is positive, which means the line has an upward trend (if you read from left to right) when x>2

[ensure the equation is a smiley or a sad one... by using #(d^2y)/dx^2#]
#(d^2y)/dx^2# = 6 (6 is positive, so this parabola is a smiley and the point(2,1) is the lowest point of the parabola), since the lowest point has a positive y coordinate, this parabola don't have common point with the x-axis

[find the common point of the equation and the y-axis]
when x = 0
y = #3x^2# - 12x + 13
= #3(0)^2# - 12(0) + 13
= 13

so the point which locates on the y-axis & the parabola is (0,13)

with these information, you can sketch a graph:
graph{3x^2 - 12x + 13 [-7.98, 23.62, -0.14, 15.66]}

reference :
#dy/dx# is the gradient of each point on a line(straight or curve)
#dy/dx# will change when x is changing
if y = #ax^p#
#dy/dx# = #a(p)x^(p-1)#

hope this help you, you can try it yourself, to be honest, sketching a graph is fun~