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**RECALL**  Quadratic Function in general form: ax 2 + bx + c where a, b, and c are real number coefficients  The sign of a determines the directions of the graph and whether there is a minimum or maximum point  The shape of the graph of a quadratic function is called a "parabola"  The parabola is symmetrical and divided in half by the axis of symmetry  The axis of symmetry is also the x – coordinate of the vertex

 Solve the quadratic equation, ax 2 + bx + c = 0, to find the x - intercepts  Use the midpoint of the x – intercepts to find the axis of symmetry  Use the axis of symmetry to find the vertex ** To Graph **  Find the y – intercept, (x = 0)  Plot the intercepts & vertex  Sketch the curve

Let's work through an example: x 2 2 x − 3 − y = 1 st : Solve the equation to find the x – intercepts (Set y = 0) x 2 2 x − 3 − y = x 2 2 x − 3 − 0 = 0 = ( x − 3 ) ( x + 1 ) x − 3 = 0 x + 1 = 0 x = 3 x = − 1 I NTERCEPT : ( 3, 0 ) I NTERCEPT : ( − 1, 0) 2 nd : Find the Axis of Symmetry 3 − 1− 1− 1− 1 Use the x – intercepts to find the x – coordinate of the Midpoint: x 1 + x 2 2 = 2 + = 1 Axis of symmetry:Axis of symmetry: x = 1 (AXIS OF SYMMETRY) x =

rd : Find the vertex using the axis of symmetry Example (continued): x 2 2 x − 3 − y = x – intercepts: ( 3, 0 ) ( − 1, 0 ) Axis of Symmetry: x = x 2 2 x − 3 − y = ( ) 2 2 − 3 − y = 1 2 − 3 − y = = −4−4 V ERTEX V ERTEX : (, ) −4−4−4−4 4 th : Find the y - intercept x 2 2 x − 3 − y = (0)(0) 2 2 (0)(0) − 3 − y = 0 0 − 3 − y = = −3−3 Y - INTERCEPT : ( 0, ) −3−3−3−3,

x y 5 th : Plot the points Example (continued): x 2 2 x − 3 − y = x – intercepts: y – intercept: Vertex: ( − 1, 0) ( 3, 0 ) ( 0, − 3) ( 1, − 4) 6 th : Sketch the curve Since the "a" coefficient is positive, we know the direction of the curve is upward.

Let's try another: −x−x 2 6 x − 8 + y = 1 st : Solve the equation ( for y = 0) −x−x 2 6 x − 8 + y = −x−x 2 6 x − = 0 = − 1(x − 2 )( x − 4 ) x − 2 = 0 x − 4 = 0 x = 2 x = 4 I NTERCEPT : ( 2, 0 ) I NTERCEPT : (4, 0) 2 nd : Find the Axis of Symmetry 2 4 Average the x-coordinates of the x – intercepts: x 1 + x 2 2 = 2 + = 3 Axis of symmetry:Axis of symmetry: x = 3 (AXIS OF SYMMETRY) x =

−8−8−8−8 −x−x rd : Find the vertex using the axis of symmetry Example (continued): 2 6 x − 8 + y = x – intercepts: ( 2, 0 ) ( 4, 0 ) Axis of Symmetry: x = −x−x 2 6 x − 8 + y = ( ) 2 6 − 8 + y = −9−9 18 − 8 + y = = 1 V ERTEX V ERTEX : (, ) 1 4 th : Find the y - intercept −x−x 2 6 x − 8 + y = (0)(0) 2 6 (0)(0) − 8 + y = 0 0 − 8 + y = = −8−8 Y - INTERCEPT : ( 0, ),

x y Since the "a" coefficient is negative, we know the direction of the curve is downward. 5 th : Plot the points Example (continued): −x−x−x−x 2 6 x − 8 + y = x – intercepts: y – intercept: Vertex: ( 2, 0) ( 4, 0 ) ( 0, − 8) ( 3, 1) 6 th : Sketch the curve

 Find the following points to graph: vertex x – intercepts y – intercept  Plot the points and sketch the curve SUMMARY  Think about the direction of your graph based on the a coefficient.