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

2. 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

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

4. 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

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

6. 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

7. -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
– x2 + 2x – 1 = 0
Parabola:
Vertex:
a = -1
b = 2
c = -1 7

8. 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

9. +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

10. 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

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