# Converting from one form to another Vertex or Root form to Standard ```Converting from one form to another
Vertex or Root form to Standard - just FOIL and multiply:
Example: y = &frac12;(x – 2)(x + 4)  y = &frac12;(x + 2x – 8)  y = &frac12;x + x – 4 (Standard form)
Standard to Vertex form – You can (a) complete the square or (b) find the axis of symmetry (x value of the vertex) and
plug x back into the equation to find the y value of the vertex.
(a) Completing the square:
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Example: y = 3x – 12x – 9
a = 3, b = -12, c = -9 (y-intercept)
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Step 1: Substitute 0 for y
0 = 3x – 12x – 9
Step 2: Move the constant
9 = 3x – 12x
Step 3: Factor-out the leading coefficient
9 + ___ = 3(x – 4x + ___ )
Step 4: Add (b/2)2 to both sides
Step 5: Factor the perfect square
Step 6: Move the constant
9 + _12_ = 3(x – 4x + _4_ ) Note: we added 3 x 4
21 = 3 (x – 2)2
y = 3 (x – 2)2 – 21
(Vertex form)
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(b) Find the axis of symmetry:
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Example: y = 3x – 12x – 9
a = 3, b = -12, c = -9 (y-intercept)
To find the vertex, use x = -b/2a: Here, x = -(-12)/2(3) = 2, so x = 2 is the Axis of Symmetry and the x-value of our
vertex.
Now, we can replace x with 2 in our equation to find the y-value of our vertex:
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y = 3(2) – 12(2) – 9 = 12 – 24 – 9 = - 21
Now, we know the value of “a” (3) and our vertex (2,-21), so we can put this in Vertex form:
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y = 3(x – 2) – 21
Pretty cool, huh?
Vertex to Roots form: When we cannot factor the Standard form to get to the Roots form, we can set y = 0 and use
the Vertex form to solve for our Roots (where the parabola crosses the x-axis):
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Example:
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y = 3(x – 2) – 21  0 = 3(x – 2) – 21
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21 = 3(x – 2)
Divide both sides by 3
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7 = (x – 2) Take the square root of both sides
&plusmn;√7 = x – 2 Add 2 to both sides and you have the Roots!
2&plusmn;√7 = x So y = 3(x – 2+ √7)(x – 2 – √7) (Root form)
Remember: Roots can be imaginary!
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Example:
y = 2(x + 1) + 2 
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0 = 2(x + 1) + 2
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-2 = 2(x + 1)
Divide both sides by 2
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-1 = (x + 1) Take the square root of both sides
Since we cannot take the square root of a negative number, our answer is the imaginary root of -1&plusmn;i
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