Download Unlike the graphical solutions used previously the use of

quadratics – completing the square
y = a(x – h)2 + k

y = ax2 +bx + c
earlier when factoring, we learned about “perfect squares”…
two numbers
that add up to
two numbers that
multiply to
2
2
e.g. x + 14x + 49 = (x + 7)(x + 7) = (x +
 14 
note:    49
2
7)2
e.g. x2 – 14x + 49 = (x – 7)(x – 7) = (x – 7)2
2
2
e.g. x + 16x + 64 = (x + 8)(x + 8) = (x +
 16 
note:    64
2
8)2
e.g. x2 – 16x + 64 = (x – 8)(x – 8) = (x – 8)2
2
2
e.g. x + 12x + 36 = (x + 6)(x + 6) = (x +
 12 
note:    36
2
6) 2
e.g. x2 – 12x + 36 = (x – 6)(x – 6) = (x – 6)2
y = x2 + 8x + 15
e.g. A
take this quadratic, put it into vertex form, find the vertex
step 1) if there is a constant common factor (“a”) for the x2 term and the x term then remove it
n/a
step 2) find the constant that you have to add and subtract to create a perfect square out of the first 2 terms
2
y = x2 + 8x + 16 – 16 + 15
8
   16
2
step 3) group the 3 terms that form the perfect square together, and
remove the subtracted constant term from inside to outside the brackets by
multiplying it with the constant common factor (“a”)
y = (x2 + 8x + 16) – 16 + 15
step 4) factor the perfect square and collect like terms
y = (x + 4)2 – 1
so, the vertex is (–4, –1)
WHY – “complete the square”? (why change a quadratic from standard to vertex form?)
a) to find the vertex of a quadratic in standard form without having to first find the zeros
b) with the vertex, we can then find the maximum or minimum value without having to graph
quadratics – completing the square
e.g. B
y = 2x2 + 12x – 3
y = ax2 +bx + c

y = a(x – h)2 + k
take this quadratic, put it into vertex form, find the vertex
step 1) if there is a constant common factor (“a”) for the x2 term and the x term then remove it
y = 2x2 + 12x – 3
y = 2(x2 + 6x) – 3
note: 2 is a constant common factor for the first 2 terms
step 2) find the constant that you have to add and subtract to create a perfect square out of the first 2 terms
2
y = 2(x2 + 6x) – 3
6
  9
2
y = 2(x2 + 6x + 9 – 9) – 3
step 3) group the 3 terms that form the perfect square together, and
remove the subtracted constant term from inside to outside the brackets by
multiplying it with the constant common factor (“a”)
y = 2(x2 + 6x + 9 – 9) – 3
y = 2(x2 + 6x + 9) – 3 – 18
note: 2(–9) = –18
step 4) factor the perfect square and collect like terms
y = 2(x + 3)2 – 21
e.g. C
y = –2x2 + 12x – 7
so, the vertex is (–3, – 21)
take this quadratic, put it into vertex form, find the vertex
step 1) if there is a constant common factor (“a”) for the x2 term and the x term then remove it
y = –2x2 + 12x – 7
y = –2(x2 – 6x) – 7
note: –2 is a constant common factor for the first 2 terms
step 2) find the constant that you have to add and subtract to create a perfect square out of the first 2 terms
2
y = –2(x – 6x) – 7
2
6
  9
2
y = –2(x2 – 6x + 9 – 9) – 7
step 3) group the 3 terms that form the perfect square together, and
remove the subtracted constant term from inside to outside the brackets by
multiplying it with the constant common factor (“a”)
y = –2(x2 – 6x + 9 – 9) – 7
y = –2(x2 – 6x + 9) – 7 + 18
note: –2(–9) = +18
step 4) factor the perfect square and collect like terms
y = –2(x – 3)2 + 11
so, the vertex is (3, 11)
quadratics – completing the square
e.g. D
y = 0.2x2 – 10x + 650
y = ax2 +bx + c

y = a(x – h)2 + k
take this quadratic, put it into vertex form, find the vertex
step 1) if there is a constant common factor (“a”) for the x2 term and the x term then remove it
y = 0.2x2 – 10x + 650
y = 0.2(x2 – 50x) + 650
note: 0.2 is a constant common factor for the first 2 terms
step 2) find the constant that you have to add and subtract to create a perfect square out of the first 2 terms
2
 50 
   625
 2 
y = 0.2(x2 – 50x) + 650
y = 0.2(x2 – 50x + 625 – 625) + 650
step 3) group the 3 terms that form the perfect square together, and
remove the subtracted constant term from inside to outside the brackets by
multiplying it with the constant common factor (“a”)
y = 0.2(x2 – 50x + 625 – 625) + 650
y = 0.2(x2 – 50x + 625) + 650 –125
note: 0.2(–625) = –125
step 4) factor the perfect square and collect like terms
y = 0.2(x – 25)2 + 525
so, the vertex is (25, 525)