What is the vertex form of #y=16x^2+14x+2#?

Given:#" "y=16x^2+14x+2#...............(1)

#color(blue)("Step 1")#

write as
#" "y=(16x^2+14x)+2#

Take the 16 outside the bracket giving:

#" "y= 16(x^2+14/16x)+2#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 2")#

This is where we start to change things but in doing so we introduce an error. This is mathematically corrected later on
At this stage it is not correct to say that it is the correct value for y

Divide what is inside the bracket by #x# giving

# " "y!= 16(x+14/16)+2#

Now halve the #14/16# inside the bracket

#" " y!= 16(x+14/32)+2" "->" " 16(x+7/16)+2#

Now square the bracket

#" "y!= 16(x+7/16)^2+2#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 3")#

In doing this we have introduced an error. This can be 'got round' like this:

Let #k# be a constant, then:

#y=16(x+7/16)^2+k+2#..........................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 4: to find the value of the correction +k")#

As both (1) and (2) equal #y# we can equate them to each other through #y#

#16x^2+14x+2" " =" " y" " = " "16(x+7/16)^2+k+2#

#16x^2+14x+2 " "=" " 16(x+7/16)^2+k+2#

Squaring the bracket

#16x^2+14x+2 " "=" " 16(x^2+14/16x+49/256)+k+2#

Multiplying the contents of the bracket by 16
we can cancel values that are the same on each side of the equals sign.

#cancel(16x^2)+cancel(14x)+cancel(2 )" "=" "cancel( 16x^2)+cancel(14x)+49/16+k+cancel(2)#

We are left with:

#0= 49/16 +k" "#so #" "k=49/16#..................(3)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 5")# Substitute (3) into (2) giving:

#y=16(x+7/16)^2+49/16+2#

So the vertex form is:

#color(red)(y=16(x+7/16)^2+ 81/16)#