The vertex form of a quadratic function is
If a > 0, then the parabola opens upward.
If a < 0, then the parabola opens downward.
Write the following quadratics in vertex form by completing the square and find the vertex.
Example 1 :
y = x2 + 2x + 4
Solution :
Using completing the square method,
y = x2 + 2⋅x⋅1 + 12 -12 + 4
y = (x+1)2 -12 + 4
y = (x+1)2 + 3
By comparing this with the vertex form of parabola, we get
(h, k) ==> (-1, 3)
Example 2 :
y = x2-6x + 3
Solution :
Using completing the square method,
y = x2 - 2⋅x⋅3 + 32 - 32 + 3
y = (x-3)2 -9 + 3
y = (x-3)2 - 6
By comparing this with the vertex form of parabola, we get
(h, k) ==> (3, -6)
Example 3 :
y = x2 - 2x
Solution :
Using completing the square method,
y = x2 - 2⋅x⋅1 + 12 - 12
y = (x-1)2 -1
By comparing this with the vertex form of parabola, we get
(h, k) ==> (1, -1)
Example 4 :
y = x2 - 3x + 1
Solution :
Using completing the square method,
y = x2 - 2⋅x⋅(3/2) + (3/2)2 - (3/2)2
y = (x-(3/2))2 - (9/4)
By comparing this with the vertex form of parabola, we get
(h, k) ==> (3/2, -9/4)
Example 5 :
y = -x2 + 2x + 8
Solution :
By factoring negative from the quadratic function, we get
y = -(x2 - 2x - 8)
y = -(x2 - 2x - 8)
y = -(x2 - 2⋅x⋅1 + 12-1-8)
y = -((x-1)2-9)
y = -(x-1)2+9
By comparing this with the vertex form of parabola, we get
(h, k) ==> (1, 9)
Example 6 :
y = (x-1)(x-3)
Solution :
y = x2-4x+3
y = x2-2⋅x⋅2+22-22+3
y = (x-2)2-4+3
y = (x-2)2-1
By comparing this with the vertex form of parabola, we get
(h, k) ==> (2, -1)
Example 7 :
y = 3(x+1)2
Solution :
y = 3(x+1)2 + 0
(h, k) ==> (-1, 0)
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