WRITING QUADRATIC FUNCTIONS IN STANDARD FORM
The standard form or vertex form of a quadratic function :
f(x) = a(x - h)2 + k
where (h, k) is the vertex.
The general form of a quadratic function :
f(x) = ax2 + bx + c
We can write the above general form of a quadratic function in standard form using the completing square method.
In each case, write the given quadratic function in standard form :
Example 1 :
f(x) = x2 - 6x + 10
Solution :
f(x) = x2 - 6x + 10
f(x) = x2 - 2(x)(3) + 10
f(x) = x2 - 2(x)(3) + 32 - 32 + 10
We know the following two algebraic identities :
(a + b)2 = a2 + 2ab + b2
(a - b)2 = a2 - 2ab + b2
So, the trinomial in the form of a2 + 2ab + b2 can be written as (a + b)2
And a2 - 2ab + b2 can be written as (a + b)2.
Therefore, x2 - 2(x)(3) + 32 can be written as (x - 3)2.
f(x) = (x - 3)2 - 32 + 10
f(x) = (x - 3)2 - 9 + 10
f(x) = (x - 3)2 + 1
Example 2 :
f(x) = x2 + 12x + 40
Solution :
f(x) = x2 + 12x + 40
f(x) = x2 + 2(x)(6) + 62 - 62 + 40
f(x) = (x + 6)2 - 62 + 40
f(x) = (x + 6)2 - 36 + 40
f(x) = (x + 6)2 - 4
Example 3 :
f(x) = x2 - 5x - 15
Solution :
f(x) = x2 - 5x - 15
Example 4 :
f(x) = x2 - 10x
Solution :
f(x) = x2 - 10x
f(x) = x2 - 2(x)(5)
f(x) = x2 - 2(x)(5) + 52 - 52
f(x) = (x - 5)2 - 52
f(x) = (x - 5)2 - 25
Example 5 :
f(x) = 2x2 + 12x + 5
Solution :
f(x) = 2x2 + 12x + 5
f(x) = 2(x2 + 6x) + 5
f(x) = 2[x2 + 2(x)(3) + 32 - 32] + 5
f(x) = 2[(x + 3)2 - 9] + 5
f(x) = 2(x + 3)2 - 18 + 5
f(x) = 2(x + 3)2 - 13
Example 6 :
f(x) = -3x2 + 9x - 4
Solution :
f(x) = -3x2 + 9x - 4
f(x) = -3(x2 - 3x) - 4
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