Solve Quadratics by Graphing ax2 +bx + c Section 10 – 3 Day 1 Solve Quadratics by Graphing ax2 +bx + c

Solve Quadratics by Graphing Remember: RIGHT SIDE of equation has to be ZERO! When solving quadratics, the following all MEAN the same thing: Find the x-values that make the y-value ZERO!! 1. Solve by Graphing 2. Solve by Factoring 4. Find the Zeros 3. Find the Solutions 6. Find the Roots 5. Find the x-intercepts 2

Graphing Quadratics cont. NOTE: – The x-intercepts are the SOLUTIONS on a graph! 3

Graphing Quadratics cont. Three Types of Solutions: – When solving quadratics, there are 3 possibilities for solutions: a. TWO Solutions b. ONE Solution c. NO Solutions 4

Example 1 Solve x2 – 6x + 5 = 0 Parabola: Vertex: a = 1 b = -6 c = 5 5

Table: 5 1 & SOLUTIONS: Example 1 cont. x (domain) y (range) y = x2 – 6x + 5 1 y = (1)2 – 6(1) + 5 y = (2)2 – 6(2) + 5 -3 2 y = (3)2 – 6(3) + 5 -4 Vertex-> 3 y = (4)2 – 6(4) + 5 -3 4 y = (5)2 – 6(5) + 5 5 1 & 5 SOLUTIONS: 6

-1 -1 – x2 + 2x – 1 = 0 Parabola: Vertex: a = -1 b = 2 c = -1 Example 2 Solve – x2 + 2x = 1 by graphing <- Make right side zero -1 -1 – x2 + 2x – 1 = 0 Parabola: Vertex: a = -1 b = 2 c = -1 7

Table: 1 SOLUTIONS: Example 2 cont. x (domain) y (range) y = – x2 + 2x – 1 -1 y = – (-1)2 + 2(-1) – 1 -4 y = – (0)2 + 2(0) – 1 -1 y = – (1)2 + 2(1) – 1 Vertex-> 1 y = – (2)2 + 2(2) – 1 -1 2 y = – (3)2 + 2(3) – 1 -4 3 1 SOLUTIONS: 8

+4x +4x x2 + 4x + 6 = 0 Parabola: Vertex: a = 1 b = 4 c = 6 Example 3 Solve x2 + 6 = – 4x by graphing <- Make right side zero +4x +4x x2 + 4x + 6 = 0 Parabola: Vertex: a = 1 b = 4 c = 6 9

Table: NONE SOLUTIONS: Example 3 cont. x (domain) y (range) y = x2 + 4x + 6 -4 y = (-4)2 + 4(-4) + 6 6 y = (-3)2 + 4(-3) + 6 3 -3 y = (-2)2 + 4(-2) + 6 2 Vertex-> -2 y = (-1)2 + 4(-1) + 6 3 -1 y = (0)2 + 4(0) + 6 6 NONE SOLUTIONS: 10

Homework Section 10-3 Day 1 Pg 647 5, 6, 9, 10, 16, 17