The document provides steps for translating a quadratic function from vertex form to standard form when a ≠ 1. It explains that the vertex form can be written as the square of a binomial. Using FOIL multiplication and the distributive property, the terms can be distributed and combined to obtain the standard form. Two examples are provided and worked through to demonstrate the process.

1. Translating Quadratic
Function from Vertex Form
into Standard Form if 𝒂 ≠ 𝟏
Mathematics 9

2. Standard From and Vertex Form of
Quadratic Function
Vertex Form
𝑦 = 𝑎 𝑥 − ℎ 2
+ 𝑘
Standard Form
𝑦 = 𝑎𝑥2
+ 𝑏𝑥 + 𝑐

3. To translate quadratic
function from vertex form back
to standard form, all we need
to do is to simplify and you
should know the following:
1. FOIL Method
2. Distributive Property

4. Vertex Form into Standard Form
 Steps for translating quadratic function from vertex back
to standard form if 𝒂 ≠ 𝟏
Step 1: Since the vertex form of quadratic equation is written in the
form of 𝑦 = 𝑎 𝑥 − ℎ 2 + 𝑘, and we have a square of binomial
which is 𝑥 − ℎ 2, then we can replace it by (𝑥 − ℎ)(𝑥 − ℎ). See
the example below.
Example 1:
𝑦 = 2 𝑥 − 3 2
+ 6
𝑦 = 2 𝑥 − 3 𝑥 − 3 + 6

5. Vertex Form into Standard Form
 Steps for translating quadratic function from vertex back
to standard form if 𝒂 ≠ 𝟏
Step 2: Now, we have two binomials and the operation is
multiplication, and to get the product of two binomials, we need to
use the FOIL method.
Example 1:
𝑦 = 2(𝑥 − 3)(𝑥 − 3) + 6
𝑭
𝑶
𝑰
𝑳
F = 𝒙 𝒙 = 𝒙 𝟐
O = 𝒙 −𝟑 = −𝟑𝒙
I = −𝟑 𝒙 = −𝟑𝒙
L = −𝟑 −𝟑 = 𝟗
𝑦 = 2(𝑥2
− 3𝑥 − 3𝑥 + 9) + 6
Do not forget to put
parenthesis because
the product of the
two binomials are still
multiplied by 2.

6. Vertex Form into Standard Form
 Steps for translating quadratic function from vertex back
to standard form if 𝒂 ≠ 𝟏
Step 3: Simplify the terms inside the parenthesis by combining like terms
Example 1:
𝑦 = 2(𝑥2
− 3𝑥 − 3𝑥 + 9) + 6
−6𝑥
𝑦 = 2(𝑥2
− 6𝑥 + 9) + 6

7. Vertex Form into Standard Form
 Steps for translating quadratic function from vertex back
to standard form if 𝒂 ≠ 𝟏
Step 4: By distributive Property, distribute 2 to all terms inside the
parenthesis.
Example 1:
𝑦 = 2(𝑥2
− 6𝑥 + 9) + 6
𝑦 = 2𝑥2
− 12𝑥 + 18 + 6
2 𝑥2 = 2𝑥2
2 −6𝑥 = −12𝑥
2 9 = 18

8. Vertex Form into Standard Form
 Steps for translating quadratic function from vertex back
to standard form if 𝒂 ≠ 𝟏
Step 5: The last step is to combine the two constant terms.
Example 1:
𝑦 = 2𝑥2
− 12𝑥 + 18 + 6
24
𝑦 = 2𝑥2
− 12𝑥 + 24
Final Answer
This is already the
standard form of the
equation
𝑦 = 2 𝑥 − 3 2 + 6

9. More Examples
Translating Vertex into Standard Form when 𝑎 ≠ 1

10. Example 1
𝒚 = − 𝒙 − 𝟖 𝟐
+ 𝟐 Quadratic in Vertex Form
𝒚 = − 𝒙 − 𝟖 𝒙 − 𝟖 + 𝟐
Square of binomial
𝑎 + 𝑏 2 = (𝑎 + 𝑏(𝑎 + 𝑏)
𝒚 = − 𝒙 𝟐 − 𝟖𝒙 − 𝟖𝒙 + 𝟔𝟒 + 𝟐 FOIL Method
𝒚 = −(𝒙 𝟐
− 𝟏𝟔𝒙 + 𝟔𝟒) + 𝟐 Simplify the result of FOIL
𝒚 = −𝒙 𝟐 + 𝟏𝟔𝒙 − 𝟔𝟒 + 𝟐 Distributive Property
𝒚 = −𝒙 𝟐 + 𝟏𝟔𝒙 − 𝟔𝟐 Combine like terms
𝒚 = −𝒙 𝟐
+ 𝟏𝟔𝒙 − 𝟔𝟐 Final Answer
Find the standard form of the function 𝒚 = − 𝒙 − 𝟖 𝟐
+ 𝟐.

11. Example 2
𝒚 = 𝟑 𝒙 + 𝟐 𝟐
− 𝟓 Quadratic in Vertex Form
𝒚 = 𝟑 𝒙 + 𝟐 𝒙 + 𝟐 − 𝟓
Square of binomial
𝑎 + 𝑏 2 = (𝑎 + 𝑏(𝑎 + 𝑏)
𝒚 = 𝟑 𝒙 𝟐 + 𝟐𝒙 + 𝟐𝒙 + 𝟒 − 𝟓 FOIL Method
𝒚 = 𝟑(𝒙 𝟐
+ 𝟒𝒙 + 𝟒) − 𝟓 Simplify the result of FOIL
𝒚 = 𝟑𝒙 𝟐 + 𝟏𝟐𝒙 + 𝟏𝟐 − 𝟓 Distributive Property
𝒚 = 𝟑𝒙 𝟐 + 𝟏𝟐𝒙 + 𝟕 Combine like terms
𝒚 = 𝟑𝒙 𝟐
+ 𝟏𝟐𝒙 + 𝟕 Final Answer
Find the standard form of the function 𝒚 = 𝟑 𝒙 + 𝟐 𝟐
− 𝟓.