Summary of 2.1 y = -x2 graph of y = x2 is reflected in the x- axis Note: Negative in front of x2 makes parabola “frown”. Reflections: y = (-x)2 graph of y = x2 is reflected in the y- axis Note: If –x is in ( ), the parabola stays “smiling”. y = Ax2 where A > 1, graph of y = x2 stretches vertically by a factor of A Note: Parabola gets “narrower” Vertical Change: y = Ax2 where 0< A < 1, graph of y = x2 shrinks vertically by a factor of A Note: Parabola gets “wider”

Summary of 2.1 y = (x - B)2 graph of y = x2 is shifted B units to the right. Horizontal Shifts: y = (x + B)2 graph of y = x2 is shifted B units to the left. Note: Think opposite direction of the sign of B. y = x2 + C graph of y = x2 is shifted C units UP. Vertical Shifts: y = x2 - C graph of y = x2 is shifted C units DOWN. Note: Follow the direction of the sign of C.

2.2 Graphing Quadratic Functions by completing the square 10/15/12

When a is positive, the parabola opens up. The vertex is minimum. Vertex Form: a quadratic equation written in the form Y = a(x –h)2 + k (h, k) is the vertex When a is positive, the parabola opens up. The vertex is minimum. When a is negative, the parabola opens down. The vertex is maximum. Graph of y = (x – 2)2 - 9: x - intercept: Where parabola crosses the x-axis. To find the x-intercept, set y = 0 and solve for x. y - intercept: Where parabola crosses the y-axis. To find the y-intercept, set x = 0 and solve for y. If equation is in standard form y = ax2 + bx + c then c is the y-intercept.

Steps for completing the square Standard form : y = x2 + bx + c y = x2 + bx ( ) + c y = (x2 + bx + 𝑏 2 2 ) + c - 𝑏 2 2 y = (x + 𝑏 2 ) 2 + d Put ( ) around x2 + bx and move c outside ( ) Take half of b and square it. Add it to the ( ) and subtract it from c. 3. Rewrite what’s in the ( ) as (x + 𝑏 2 )2 and simplify c - 𝑏 2 2 Note: if y = x2 – bx + c Then y = (x - 𝑏 2 )2 + d

Example 1 y = x2 + 6x ( ) + 5 y = (x2 + 6x + 9) + 5 - 9 Rewrite y = x2 + 6x + 5 in Vertex Form and describe the transformation. Determine the x-int and y int. (if it exists) and vertex. Determine if the vertex is a maximum or a minimum. a.) Rewrite in vertex form and describe the transformation. y = x2 + 6x ( ) + 5 Put ( ) around x2 + 6x and move +5 outside ( ) Take half of 6 and square it. Add 9 to the ( ) and subtract 9 from 5. 3. Rewrite what’s in the ( ) as (x + 3)2 y = (x2 + 6x + 9) + 5 - 9 y = (x + 3) 2 - 4 Graph of x2 shifts 3 units to the left and 4 units down.

Example 1 Since a is positive 1, the vertex is minimum. b.) Determine the x and y intercepts and vertex x –int. set y = 0 and solve for x. y –int. set x = 0 and solve for y. 0 = (x + 3) 2 - 4 +4 +4 4 = (x + 3)2 ±2 = x + 3 2 = x + 3 and -2 = x + 3 -1 = x -5 = x y = (0 + 3) 2 - 4 y = (3)2 - 4 y = 9 - 4 y = 5 Or since the original problem is written in standard form, y –int is the c term which is 5. y = x2 +6x + 5 Vertex (h, k) = (-3, -4) c.) Minimum or Maximum? Since a is positive 1, the vertex is minimum.

Example 2 y = x2 - 6x ( ) + 10 y = (x2 - 6x + 9) + 10 - 9 Rewrite y = x2 - 6x + 10 in Vertex Form and describe the transformation. Determine the x-int and y int. (if it exists) and vertex. Determine if the vertex is a maximum or a minimum. a.) Rewrite in Vertex Form and describe the transformation. y = x2 - 6x ( ) + 10 Put ( ) around x2 - 6x and move +10 outside ( ) Take half of 6 and square it. Add 9 to the ( ) and subtract 9 from 10. 3. Rewrite what’s in the ( ) as (x - 3)2 y = (x2 - 6x + 9) + 10 - 9 y = (x - 3) 2 + 1 Graph of x2 shifts 3 units to the right and 1 units up.

Example 2 Since a is positive 1, the vertex is a minimum. y –int. b.) Determine the x and y intercepts and vertex x –int. set y = 0 and solve for x. y –int. y = x2 -6x + 10 Since the original problem is written in standard form, y –int is the c term which is 10. 0 = (x - 3) 2 + 1 -1 -1 -1 = (x - 3)2 Since you cannot take the square root of a negative number, there is no x- intercept Vertex (h, k) = (3, 1) c.) Minimum or Maximum Since a is positive 1, the vertex is a minimum.

Example 3 y = 2x2 - 4x + 6 2 2 2 2 𝑦 2 = x2 - 2x ( ) + 3 Rewrite y = 2x2 - 4x + 6 in Vertex Form and describe the transformation. Determine the x-int and y int. (if it exists) and vertex. Determine if the vertex is a maximum or a minum. a.) Rewrite y = x2 - 6x + 10 in Vertex Form and describe the transformation. y = 2x2 - 4x + 6 Divide everything by 2. Put ( ) around x2 - 2x and move +3 outside ( ) Take half of 2 and square it. Add 1 to the ( ) and subtract 1 from 3. 4. Rewrite what’s in the ( ) as (x - 1)2 5. Multiply both sides by 2. 2 2 2 2 𝑦 2 = x2 - 2x ( ) + 3 𝑦 2 = (x2 - 2x +1) + 3 - 1 𝑦 2 = (x - 1) 2 + 2 𝑦 = 2(x - 1) 2 + 4 Graph of x2 stretches vertically by factor of 2 and shifts 1 unit to the right and 4 units up.

Example 3 Since a is positive 2, the vertex is a minimum. y –int. b.) Determine the x and y intercepts and vertex x –int. set y = 0 and solve for x. y –int. y = 2x2 -4x + 6 Since the original problem is written in standard form, y –int is the c term which is 6. 0 = 2(x - 1) 2 + 4 -4 -4 -4 = 2(x - 1)2 Divide by 2 -2 = (x - 1)2 Since you cannot take the square root of a negative number, there is no x- intercept Vertex (h, k) = (1, 4) c.) Minimum or Maximum Since a is positive 2, the vertex is a minimum.