1. Concept 24 Essential Question/Topic: I can change a quadratic from standard form into vertex form.

2. 24.1 I can change quadratics from standard from into vertex form. Convert from standard form to vertex form. 1.“a” is the coefficient of the x 2 term. 2.Use the formula to find h, -b/2a (the same as the AOS formula!) 3.Substitute the value you found for h into the original equation and solve for k. Standard Form  Vertex Form Y = ax 2 + bx + c  y = a(x – h) 2 + k Ex1 Y = 8x 2 – 16x + 27 We know a, b, and c. We want a, h, and k. a = 8 h = -(-16)/2(8) h = 1 k = (y = 8(x) 2 – 16(x) + 27) k = (y = 8(h) 2 – 16(h) + 27) k = (y = 8(1) 2 – 16(1) + 27) = 19 Now, plug all of the values in! y = a(x – h) 2 + k y = 8(x – 1) 2 + 19 Questions/Central IdeasNotes/Key Details

3. 24.1 I can change quadratics from standard from into vertex form. Convert from standard form to vertex form. 1.“a” is the coefficient of the x 2 term. 2.Use the formula to find h, -b/2a (the same as the AOS formula!) 3.Substitute the value you found for h into the original equation and solve for k. Standard Form  Vertex Form Y = ax 2 + bx + c  y = a(x – h) 2 + k Ex2 Y = 5x 2 – 40x + 67 We know a, b, and c. We want a, h, and k. a = 5 h = -(-40)/2(5) = 4 h = 4 k = (y = 5(x) 2 – 40(x) + 67) k = (y = 5(h) 2 – 40(h) + 67) k = (y = 5(4) 2 – 40(4) + 67) = -13 Now, plug all of the values in! y = a(x – h) 2 + k y = 5(x – 4) 2 – 13 Questions/Central IdeasNotes/Key Details

4. 24.1 I can identify the vertex and axis of symmetry from vertex form and graph the equation. Graphing a parabola when in vertex form. 1.Find the Axis Of Symmetry. Set the parenthetical equal to zero and solve for x. 2.Find the vertex (lowest or highest point of the parabola) by plugging your AOS in for x. Vertex Form  y = a(x – h) 2 + k Ex1 y = 2(x – 2) 2 + 1 (x – 2) = 0 x = 2   Axis of symmetry    y = 2(x – 2) 2 + 1 y = 2(2 – 2) 2 + 1 y = 2(0) 2 + 1 y = 1 Vertex = (2,1) Questions/Central IdeasNotes/Key Details

5. 24.1 I can identify the vertex and axis of symmetry from vertex form and graph the equation. 3.Find more points to your parabola by choosing two x- values on each side of the vertex. 4.Plug each one of the x-values into your initial equation and solve to get a complete coordinate. Vertex = (2,1) Initial equation: y = 2(x – 2) 2 + 1 y = 2(1 – 2) 2 + 1 y = 2(-1) 2 + 1 y = 2(1) + 1 y = 3 Your coordinate is (1,3). Since it’s a parabola, when you plug in a 3 for x, you will also get out 3. Same with your x-values of 0 and 4, they will have the same y-values. Questions/Central IdeasNotes/Key Details X01234 Y1 X01234 Y93139

6. 24.1 I can identify the vertex and axis of symmetry from vertex form and graph the equation. 1.Plot the points to graph your parabola! Questions/Central IdeasNotes/Key Details X01234 Y93139