To convert standard form to vertex form, we may follow the different ways.
From the standard form of the equation, y = ax^{2} + bx + c
Step 1 :
Factor the leading coefficient from x^{2} terms and x term.
Step 2 :
Write the coefficient of x as a multiple of 2.
Step 3 :
Comparing x^{2} terms and x term with the algebraic identity a^{2} + 2ab or a^{2} - 2ab, the missing term will be b^{2}.
Step 4 :
So, add b^{2}. In order to balance our work, we have to subtract b^{2}.
Step 5 :
The first three terms will match with any one of the forms as a^{2} + 2ab + b^{2} or a^{2} - 2ab + b^{2}, we can compress it as using the algebraic identities (a+b)^{2} or (a-b)^{2}.
Step 6 :
After combining the like terms, we get the vertex form
y = a(x - h)^{2} + k
(or)
y = -a(x - h)^{2} + k
From the standard form of the equation, y = ax^{2} + bx + c
(i) Factor the leading coefficient from x^{2 }term and x term.
(ii) Take half of the coefficient of x and write it as (x - a)^{2} or (x + a)^{2. }Here a is half the coefficient of x.
Write each function in vertex form :
Problem 1 :
y = x^{2} - 6x + 3
Solution :
y = x^{2} - 6x + 3
The leading coefficient is 1. So, let us write the coefficient of x as multiple of 2.
y = x^{2} - 2x(3) + 3
Comparing the first two terms x^{2} - 2x(3), looks like a^{2} - 2ab. In order to complete the formula, we need b^{2}. Then add b^{2}, to balance our work, we need to write -b^{2}.
y = x^{2} - 2x(3) + 3^{2} - 3^{2} + 3
Now combining a^{2} - 2ab + b^{2}, we get (a - b)^{2}.
y = (x - 3)^{2} - 3^{2} + 3
y = (x - 3)^{2} - 9 + 3
y = (x - 3)^{2} - 6
Vertex of the parabola is (3, -6).
Problem 2 :
y = x^{2} + 2x + 7
Solution :
y = x^{2} + 2x + 7
The leading coefficient is 1. So, let us write the coefficient of x as multiple of 2.
y = x^{2} + 2x(1) + 7
y = x^{2} + 2x(1) + 1^{2} - 1^{2} + 7
y = (x + 1)^{2} - 1 + 7
y = (x + 1)^{2} + 6
Vertex of the parabola is (-1, 6).
Problem 3 :
y = x^{2} + 9x + 7
Solution :
y = x^{2} + 9x + 7
The leading coefficient is 1. So, let us write the coefficient of x as multiple of 2. Here the coefficient of x is 9, which is odd number. In order to write it as multiple of 2, we multiply and divide by 2.
So, the vertex of the parabola is (-9/2, -49/4).
Problem 4 :
y = -3x^{2} + 12x - 10
Solution :
y = -3x^{2} + 12x - 10
Factoring -3 from x^{2} term and x term, we get
y = -3[x^{2} - 4x] - 10
y = -3[x^{2} - 2 x (2) + 2^{2} - 2^{2}] - 10
y = -3[(x - 2)^{2} - 4] - 10
y = -3(x - 2)^{2} + 12 - 10
y = -3(x - 2)^{2} + 2
The vertex of the parabola is (2, 2).
Problem 5 :
y = 3x^{2} + 10x
Solution :
y = 3x^{2} + 10x
Factoring -3 from x^{2} term and x term, we get
y = 3[x^{2} + (10/3)x]
y = 3[x^{2} + 2 ⋅ x ⋅ (5/6) + (5/6)^{2} - (5/6)^{2}]
y = 3[(x + (5/6))^{2} - (5/6)^{2}]
y = 3[(x + (5/6))^{2} - (25/36)]
y = 3(x + (5/6))^{2} - (25/12)
The vertex of the parabola is (-5/6, -25/12).
Problem 6 :
y = x^{2} - 12x + 36
Solution :
y = x^{2} - 12x + 36
y = x^{2} - 2 x (6) + 6^{2} - 6^{2} + 36
y = x^{2} - 2 x (6) + 6^{2} - 36 + 36
y = (x - 6)^{2}
So, the vertex of the parabola is (6, 0).
Problem 7 :
y = -4x^{2} - 24x - 15
Solution :
y = -4x^{2} - 24x - 15
y = -4[x^{2} + 6x] - 15
y = -4[x^{2} + 2 x (3) + 3^{2} - 3^{2}] - 15
y = -4[(x + 3)^{2} - 9] - 15
y = -4(x + 3)^{2} + 36 - 15
y = -4(x + 3)^{2} + 21
So, the vertex of the parabola is (-3, 21).
Problem 8 :
y = -x^{2} - 4x - 1
Solution :
y = -x^{2} - 4x - 1
y = -[x^{2} + 4x] - 1
y = -[x^{2} + 2x(2) + 2^{2} - 2^{2}] - 1
y = -[(x + 2)^{2} - 4] - 1
y = -(x + 2)^{2} + 4 - 1
y = -(x + 2)^{2} + 3
So, the vertex of the parabola is (-2, 3).
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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