Given one of the forms of a quadratic function, convert to the other two forms.
(ii) Factored form
(i) Converting into vertex form :
Coefficient of x 2 is 1. So, don't have to factories anything.
Write the coefficient of x as multiple of 2.
y = x 2 − 2 ⋅x ⋅3 + 3 2 - 3 2 - 7
y = (x - 3) 2 - 3 2 - 7
y = (x - 3) 2 - 9 - 7
Vertex (h, k) is (3, -16).
(i) Converting into factored form :
Decomposing -7, we get 1 (-7).
y = x 2 + 1x - 7x − 7
y = x(x + 1) - 7(x + 1)
So, -1 and 7 are x-intercepts.
(i) Standard form
(ii) Factored form
(i) Converting into standard form :
Here (x-1) 2 looks like (a - b) 2 .
(a - b) 2 = a 2 - 2ab + b 2
(x − 1) 2 = x 2 - 2x(1) + 1 2
(x − 1) 2 = x 2 - 2x - 3
y = 3(x 2 - 2x + 1) - 3
y = 3x 2 - 6x + 3 - 3
(ii) Converting into factored form :
Factor 3x, we get
So, 0 and 2 are x-intercepts.
(ii) Standard form:
(i) Converting into vertex form :
y = 2(x 2 + 5x - 3x - 15)
y = 2(x 2 + 2x - 15)
y = 2(x 2 + 2 ⋅ x ⋅1 + 1 2 - 1 2 - 15)
y = 2[(x + 1) 2 - 1 2 - 15]
So, vertex is (-1, -32).
(ii) Converting into standard form :
y = 2(x 2 + 5x - 3x - 15)
y = 2(x 2 + 2x - 15)
Distributing 2, we get
y = 2x 2 + 4x - 30
y = 2x 2 + 8x + 10
(ii) Factored form
(i) Converting into vertex form :
y = 2x 2 + 8x - 10
y = 2( x 2 + 4x - 5)
y = 2( x 2 + 2 ⋅x ⋅2 + 2 2 - 2 2 - 5)
So, the vertex is (-2, -18).
(ii) Converting into factored form :
y = 2x 2 + 8x - 10
y = 2(x - 5 2 + 4x - 5)
(i) Standard form
(ii) Factored form
(i) Converting into standard form :
Using the algebraic identity (a + b) 2 , we can find the expansion of (x + 1) 2
(x + 1) 2 = x 2 + 2 ⋅x ⋅1 + 1 2
(x + 1) 2 = x 2 + 2 x + 1
y = 2( x 2 + 2 x + 1 ) − 8
y = 2 x 2 + 4 x + 2 − 8
y = 2 x 2 + 4 x − 6
(ii) Converting into factored form :
y = 2 x 2 + 4 x − 6
Factoring 2, we get
y = 2( x 2 + 2 x − 3)
(ii) Standard form:
(i) Converting into vertex form :
y = −4(2x 2 - 5x + 2x - 5)
y = −4(2x 2 - 3x - 5)
y = −4(2x 2 - 3x - 5)