Algebra 2 HW #31: Converting between Standard and Vertex Forms Name_______________________________ 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 π¦ = 2π₯ 2 + 4π₯ β 7 from standard form into vertex form and then graph. π = πππ + ππ β π π¦ + 7 = 2π₯ 2 + 4π₯ Standard Form π¦ + 7 = 2(π₯ 2 + 2π₯) Note ( ) = ( ) = 1 π 2 2 π¦ + 7 + 2(1) = 2(π₯ 2 + 2π₯ + (1)) π¦ + 9 = 2(π₯ + 1)2 π = π(π + π)π β π Vertex form Graph: 1. πΏππππ‘π π£πππ‘ππ₯ (β1, β9) 2. πΏππππ‘π π¦ β πππ‘ππππππ‘ ππ¦ π ππ‘π‘πππ π₯ = 0 π¦ = 2(0 + 1)2 β 9 π¦ = β7; π¦ β πππ‘ππππππ‘ πππππ‘ππ ππ‘ (0, β7) 3. πΏππππ‘π πππππππ‘πππ πππππ‘ ππ π¦ β πππ‘ππππππ‘ (β2, β7) y ο±ο± ο±ο° οΉ οΈ ο· οΆ ο΅ ο΄ ο³ ο² ο± οο±ο± οο±ο° οοΉ οοΈ οο· οοΆ οο΅ οο΄ οο³ οο² οο± οο± οο² οο³ οο΄ οο΅ οοΆ οο· οοΈ οοΉ οο±ο° οο±ο± x ο± ο² ο³ ο΄ ο΅ οΆ ο· οΈ οΉ ο±ο° ο±ο± 2 2 2 You Try: Show all work Convert to Vertex Form and then graph using Vertex Form Graphing 1. π¦ = π₯ 2 + 4π₯ β 12 ο±ο΄ y ο±ο³ ο±ο² ο±ο± ο±ο° οΉ οΈ ο· οΆ ο΅ ο΄ ο³ ο² ο± x οο±ο΄ οο±ο³ οο±ο² οο±ο± οο±ο° οοΉ οοΈ οο· οοΆ οο΅ οο΄ οο³ οο² οο±οο± ο± ο² ο³ ο΄ ο΅ οΆ ο· οΈ οΉ ο±ο° ο±ο± ο±ο² ο±ο³ ο±ο΄ οο² οο³ οο΄ οο΅ οοΆ οο· οοΈ οοΉ οο±ο° οο±ο± οο±ο² οο±ο³ οο±ο΄ 2. π¦ = β3π₯ 2 β 6π₯ + 9 ο±ο΄ y ο±ο³ ο±ο² ο±ο± ο±ο° οΉ οΈ ο· οΆ ο΅ ο΄ ο³ ο² ο± x οο±ο΄ οο±ο³ οο±ο² οο±ο± οο±ο° οοΉ οοΈ οο· οοΆ οο΅ οο΄ οο³ οο² οο±οο± ο± ο² ο³ ο΄ ο΅ οΆ ο· οΈ οΉ ο±ο° ο±ο± ο±ο² ο±ο³ ο±ο΄ οΈ οΉ ο±ο° ο±ο± ο±ο² ο±ο³ ο±ο΄ οο² οο³ οο΄ οο΅ οοΆ οο· οοΈ οοΉ οο±ο° οο±ο± οο±ο² οο±ο³ οο±ο΄ 3. π¦ = 2π₯ 2 β 6π₯ β 8 ο±ο΄ y ο±ο³ ο±ο² ο±ο± ο±ο° οΉ οΈ ο· οΆ ο΅ ο΄ ο³ ο² ο± οο±ο΄ οο±ο³ οο±ο² οο±ο± οο±ο° οοΉ οοΈ οο· οοΆ οο΅ οο΄ οο³ οο² οο±οο± οο² οο³ οο΄ οο΅ οοΆ οο· οοΈ οοΉ οο±ο° οο±ο± οο±ο² οο±ο³ οο±ο΄ x ο± ο² ο³ ο΄ ο΅ οΆ ο· To convert from vertex form to standard form, you must simplify the equation remembering to use PEMDAS. Example: Convert π¦ = 2(π₯ + 2)2 β 2 from vertex form into standard form and then graph. π = π(π + π)π β π π¦ = 2(π₯ + 2)(π₯ + 2) β 2 π¦ = 2(π₯ 2 + 4π₯ + 4) β 2 π¦ = 2π₯ 2 + 8π₯ + 8 β 2 Vertex Form π = πππ + ππ + π Standard Form Graph: Use 5 point method 1. L.O.S.: π₯ = βπ 2π β8 = 2(2) = β2 2. Vertex: π¦ = 2(β2)2 + 8(β2) + 6 π¦ = β2 ππππ‘ππ₯: (β2, β2) 3. X-intercepts: 0 = 2π₯ 2 + 8π₯ + 6 0 = 2(π₯ 2 + 4π₯ + 3) 0 = 2(π₯ + 3)(π₯ + 1) π₯ = β3 πππ π₯ = β1 π₯ β πππ‘ππππππ‘π : (β3,0) πππ (β1,0) 4. Y-intercept: π¦ = 2(0)2 + 8(0) + 6 π¦=6 π¦ β πππ‘ππππππ‘: (0,6) 5. Reflection Point of y-intercept: (-4, 6) ο±ο΄ y ο±ο³ ο±ο² ο±ο± ο±ο° οΉ οΈ ο· οΆ ο΅ ο΄ ο³ ο² ο± οοΉ οοΈ οο· οοΆ οο΅ οο΄ οο³ οο² οο± οο± οο² οο³ οο΄ οο΅ οοΆ οο· οοΈ οοΉ οο±ο° οο±ο± οο±ο² οο±ο³ οο±ο΄ x ο± ο² ο³ ο΄ ο΅ οΆ ο· οΈ οΉ You Try: Show all work Convert to Standard Form and then graph using 5 point method 1. π¦ = (π₯ β 6)2 β 16 ο±ο΄ y ο±ο³ ο±ο² ο±ο± ο±ο° οΉ οΈ ο· οΆ ο΅ ο΄ ο³ ο² ο± x οο±ο΄ οο±ο³ οο±ο² οο±ο± οο±ο° οοΉ οοΈ οο· οοΆ οο΅ οο΄ οο³ οο² οο±οο± ο± ο² ο³ ο΄ ο΅ οΆ ο· οΈ οΉ ο±ο° ο±ο± ο±ο² ο±ο³ ο±ο΄ οο² οο³ οο΄ οο΅ οοΆ οο· οοΈ οοΉ οο±ο° οο±ο± οο±ο² οο±ο³ οο±ο΄ 2. π¦ = β(π₯ + 3)2 + 16 ο±ο΄ y ο±ο³ ο±ο² ο±ο± ο±ο° οΉ οΈ ο· οΆ ο΅ ο΄ ο³ ο² ο± x οο±ο΄ οο±ο³ οο±ο² οο±ο± οο±ο° οοΉ οοΈ οο· οοΆ οο΅ οο΄ οο³ οο² οο±οο± ο± ο² ο³ ο΄ ο΅ οΆ ο· οΈ οΉ ο±ο° ο±ο± ο±ο² ο±ο³ ο±ο΄ οΈ οΉ ο±ο° ο±ο± ο±ο² ο±ο³ ο±ο΄ οο² οο³ οο΄ οο΅ οοΆ οο· οοΈ οοΉ οο±ο° οο±ο± οο±ο² οο±ο³ οο±ο΄ 3. π¦ = 2(π₯ + 2)2 β 8 ο±ο΄ y ο±ο³ ο±ο² ο±ο± ο±ο° οΉ οΈ ο· οΆ ο΅ ο΄ ο³ ο² ο± οο±ο΄ οο±ο³ οο±ο² οο±ο± οο±ο° οοΉ οοΈ οο· οοΆ οο΅ οο΄ οο³ οο² οο±οο± οο² οο³ οο΄ οο΅ οοΆ οο· οοΈ οοΉ οο±ο° οο±ο± οο±ο² οο±ο³ οο±ο΄ x ο± ο² ο³ ο΄ ο΅ οΆ ο·
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