Converting Between Standard and Vertex Form

Algebra 2
HW #31: Converting between Standard and Vertex Forms
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To convert from standard form to vertex form, you must complete the square.
Steps:
1. Isolate all x-terms on one side of the equation. The other side of the equation should have y and any constants.
2. Complete the square on the x-terms. Remember to add an equivalent value to the other side.
3. Solve for y
4. Now graph the equation in vertex form (locate vertex, y-intercept and reflection point)
Example:
Convert 𝑦 = 2π‘₯ 2 + 4π‘₯ βˆ’ 7 from standard form into vertex form and then graph.
π’š = πŸπ’™πŸ + πŸ’π’™ βˆ’ πŸ•
𝑦 + 7 = 2π‘₯ 2 + 4π‘₯
Standard Form
𝑦 + 7 = 2(π‘₯ 2 + 2π‘₯)
Note ( ) = ( ) = 1
𝑏 2
2
𝑦 + 7 + 2(1) = 2(π‘₯ 2 + 2π‘₯ + (1))
𝑦 + 9 = 2(π‘₯ + 1)2
π’š = 𝟐(𝒙 + 𝟏)𝟐 βˆ’ πŸ—
Vertex form
Graph:
1. πΏπ‘œπ‘π‘Žπ‘‘π‘’ π‘£π‘’π‘Ÿπ‘‘π‘’π‘₯ (βˆ’1, βˆ’9)
2. πΏπ‘œπ‘π‘Žπ‘‘π‘’ 𝑦 βˆ’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘’π‘π‘‘ 𝑏𝑦 𝑠𝑒𝑑𝑑𝑖𝑛𝑔 π‘₯ = 0
𝑦 = 2(0 + 1)2 βˆ’ 9
𝑦 = βˆ’7; 𝑦 βˆ’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘’π‘π‘‘ π‘™π‘œπ‘π‘Žπ‘‘π‘’π‘‘ π‘Žπ‘‘ (0, βˆ’7)
3. πΏπ‘œπ‘π‘Žπ‘‘π‘’ π‘Ÿπ‘’π‘“π‘™π‘’π‘π‘‘π‘–π‘œπ‘› π‘π‘œπ‘–π‘›π‘‘ π‘œπ‘“ 𝑦 βˆ’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘’π‘π‘‘ (βˆ’2, βˆ’7)
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You Try: Show all work
Convert to Vertex Form and then graph using Vertex Form Graphing
1. 𝑦 = π‘₯ 2 + 4π‘₯ βˆ’ 12
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2. 𝑦 = βˆ’3π‘₯ 2 βˆ’ 6π‘₯ + 9
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3. 𝑦 = 2π‘₯ 2 βˆ’ 6π‘₯ βˆ’ 8
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To convert from vertex form to standard form, you must simplify the equation remembering to use PEMDAS.
Example:
Convert 𝑦 = 2(π‘₯ + 2)2 βˆ’ 2 from vertex form into standard form and then graph.
π’š = 𝟐(𝒙 + 𝟐)𝟐 βˆ’ 𝟐
𝑦 = 2(π‘₯ + 2)(π‘₯ + 2) βˆ’ 2
𝑦 = 2(π‘₯ 2 + 4π‘₯ + 4) βˆ’ 2
𝑦 = 2π‘₯ 2 + 8π‘₯ + 8 βˆ’ 2
Vertex Form
π’š = πŸπ’™πŸ + πŸ–π’™ + πŸ”
Standard Form
Graph: Use 5 point method
1. L.O.S.: π‘₯ =
βˆ’π‘
2π‘Ž
βˆ’8
= 2(2) = βˆ’2
2. Vertex:
𝑦 = 2(βˆ’2)2 + 8(βˆ’2) + 6
𝑦 = βˆ’2
π‘‰π‘’π‘Ÿπ‘‘π‘’π‘₯: (βˆ’2, βˆ’2)
3. X-intercepts:
0 = 2π‘₯ 2 + 8π‘₯ + 6
0 = 2(π‘₯ 2 + 4π‘₯ + 3)
0 = 2(π‘₯ + 3)(π‘₯ + 1)
π‘₯ = βˆ’3 π‘Žπ‘›π‘‘ π‘₯ = βˆ’1
π‘₯ βˆ’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘’π‘π‘‘π‘ : (βˆ’3,0) π‘Žπ‘›π‘‘ (βˆ’1,0)
4. Y-intercept:
𝑦 = 2(0)2 + 8(0) + 6
𝑦=6
𝑦 βˆ’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘’π‘π‘‘: (0,6)
5. Reflection Point of y-intercept: (-4, 6)
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You Try: Show all work
Convert to Standard Form and then graph using 5 point method
1. 𝑦 = (π‘₯ βˆ’ 6)2 βˆ’ 16
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2. 𝑦 = βˆ’(π‘₯ + 3)2 + 16
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3.
𝑦 = 2(π‘₯ + 2)2 βˆ’ 8
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