General Form and Completing the Square

General Form and Completing the Square

Transformation Form
(y-k)=(x-h)2 / ß à
(rearrange the equation) / Standard Form
y=a(x-h)2+k
Transformation Form
or
Standard Form / à
(isolate for y by expanding and collecting like term) / General Form
y=ax2+bx+c
Transformation Form
or
Standard Form / ß
(“complete the square”) / General Form
y=ax2+bx+c

Transformational form

·  Easy to know the vertex

·  Easy to know the stretch

·  Cannot be entered into a graphing calculator to graph

Standard form

·  Vertex can be found

·  Stretch can be found

·  Can be entered into a graphing calculator to graph

General form

·  Vertex is not easy to find

·  Stretch can be found

·  Easy to know the y intercept

·  Can be entered into a graphing calculator to graph

How to Complete the Square

Recall:

Ex: (x-4)2= (x-4)(x-4)= x2-4x-4x+16= x2-8x+16

Ex: (x+5)2= (x+5)(x+5)= x2+5x+5x+25= x2+10x+25

**This is the basic concept behind completing the square**

Examples where a=1 and c=0
Given y=x2+10x
·  Note a=1 b=10 c=0
·  Take half of b and then square it
·  10/2= 5 and 52=25
·  This value needs to be added to x2+10x to complete the square, BUT what is done to one side of an equation must also be done to the other
·  y+25= x2+10x+25
·  Because 25 is the result of taking half of b and squaring it, the equation can be written as
·  y+25 = (x+5)2 in transformation form or y=(x+5)2-25 in standard form
Given y=x2-4x
·  Note a=1 b=-4 c=0
·  Take half of b and then square it
·  -4/2= -2 and (-2)2=4
·  This value needs to be added to x2-4x to complete the square, BUT what is done to one side of an equation must also be done to the other
·  y+4= x2-4x+4
·  Because 4 is the result of taking half of b and squaring it, the equation can be written as
·  y+4= (x-2)2 in transformation form or y=(x-2)2-4 in standard form
Examples where a=1 and c0
Given y=x2+2x+6
·  Note a=1 b=2 c=6
·  Subtract c from both sides
·  y-6=x2+2x
·  Take half of b and then square it
·  2/2= 1 and (1)2=1
·  This value needs to be added to x2+2x to complete the square, BUT what is done to one side of an equation must also be done to the other
·  y-6+1= x2+2x+1
·  Now we can collect like terms on each side of the equation
·  y-5= x2+2x+1
·  Because 1 is the result of taking half of b and squaring it, the equation can be written as
·  y-5= (x+1)2 in transformation form or y=(x+1)2+5 in standard form
Given y=x2-8x-10
·  Note a=1 b=-8 c=-10
·  Subtract c from both sides
·  y+10=x2-8x
·  Take half of b and then square it
·  -8/2= -4 and (-4)2=16
·  This value needs to be added to x2-8x to complete the square, BUT what is done to one side of an equation must also be done to the other
·  y+10+16= x2-8x+16
·  Now we can collect like terms on each side of the equation
·  y+26= x2-8x+16
·  Because 16 is the result of taking half of b and squaring it, the equation can be written as
·  y+26= (x-4)2 in transformation form or y=(x-4)2-26 in standard form
Examples where a1 and c0
Given y=½x2+10x-7
METHOD 1
·  Note a= ½ b=10 c=-7
·  Subtract c from both sides
·  y+7= ½x2+10x
·  Factor out ‘a’ (the goal is to get a=1)
·  y+7= ½(x2+20x)
·  Take half of b and then square it
·  20/2= 10 and (10)2=100
·  This value needs to be added INSIDE THE BRACKETS of ½(x2+20x) to complete the square, BUT what is done to one side of an equation must also be done to the other
·  CAREFULL: since ½ is being multiplied to everything inside the brackets, really we are only adding 50 to both sides since ½(100)=50
·  y+7+50= ½(x2+20x+100)
·  Now we can collect like terms on each side of the equation
·  y+57= ½(x2+20x+100)
·  Because 100 is the result of taking half of b and squaring it, the equation can be written as
·  y+57= ½(x+10)2
·  So this looks like 2(y+57)=(x+10)2 in transformation form
·  Or y= ½(x-+10)2-57 in standard form / METHOD 2
·  Note a= ½ b=10 c=-7
·  Multiply both sides of the equation by the reciprocal of the ‘a’
·  REMEMBER ALL TERMS ARE EFFECTED
·  2y=x2+20x-14
·  Subtract c from both sides
·  2y+14= x2+20x
·  Take half of b and then square it
·  20/2= 10 and (10)2=100
·  This value needs to be added to x2+20x to complete the square, BUT what is done to one side of an equation must also be done to the other
·  2y+14+100= (x2+20x+100)
·  Now we can collect like terms on each side of the equation
·  2y+114= (x2+20x+100)
·  Because 100 is the result of taking half of b and squaring it, the equation can be written as
·  2y+114=(x+10)2
·  We must factor out the number in front of y so we get
·  2(y+57)=(x+10)2
·  So this looks like 2(y+57)=(x+10)2 in transformation form
·  Or y= ½(x-+10)2-57 in standard form

Complete the square for the following questions.

Give your answer in both transformational and standard form.

1a) y=x2-4x b) y=x2+10x c) y=x2-18x d) y=x2+6x
e) y=x2-8x f) y=x2+8x g) y=x2-50x h) y=x2+88x
i) y=x2-20x j) y=x2+18x k) y=x2-100x l) y=x2+80x
m) y=x2-16x n) y=x2+26x o) y=x2-22x p) y=x2+122x
2a) y=x2-2x+4 b) y=x2+8x+10 c) y=x2-2x-9 d) y=x2+16x+10
e) y=x2-8x-3 f) y=x2+18x+1 g) y=x2-6x+4 h) y=x2+8x-7
i) y=x2-10x+3 j) y=x2+82x-1 k) y=x2-10x-1 l) y=x2+22x-3
m) y=x2-6x-10 n) y=x2+20x+9 o) y=x2-2x+100 p) y=x2+12x+6
3a) y=2x2-4x+6 b) y=½x2+10+1 c) y=3x2-9x-3 d) y=x2+6x-3
e) y=2x2-8x-10 f) y=½x2+8x-3 g) y=3x2-30x+9 h) y=x2+2x+1
i) y=3x2-12x+9 j) y=x2+2x +2 k) y=2x2-100x-2 l) y=½x2+8x-5
m) y=3x2-6x-12 n) y=x2+6x-3 o) y=2x2-22x+60 p) y=½x2+12x+3