6.7 Graph and Solving Quadratic Inequalities
Jun 17, 2015
6.7 Graph and Solving Quadratic Inequalities
Method ofGraph
sketching
Forms of Quadratic InequalitiesForms of Quadratic Inequalitiesy<ax2+bx+c y>ax2+bx+cy≤ax2+bx+c y≥ax2+bx+c
Graphs will look like a parabola with a solid or dotted line and a shaded section.
The graph could be shaded inside the parabola or outside.
Steps for graphingSteps for graphing
1. Sketch the parabola y=ax2+bx+c
(dotted line for < or >, solid line for ≤ or ≥)
** remember to use 5 points for the graph!
2. Choose a test point and see whether it is a solution of the inequality.
3. Shade the appropriate region.
(if the point is a solution, shade where the point is, if it’s not a solution, shade the other region)
Example:Graph y ≤ x2+6x- 4
3)1(2
6
2
a
bx* Vertex: (-3,-13)
* Opens up, solid line
134189
4)3(6)3( 2
y 9- 5-
12- 4-
13- 3-
12- 2-
9- 1-
yx
•Test Point: (0,0)
0≤02+6(0)-4
0≤-4 So, shade where the point is NOT!
Test point
Graph: y>-x2+4x-3
* Opens down, dotted line.
* Vertex: (2,1)2
)1(2
4
2
a
bx
1384
3)2(4)2(1 2
y
y
* Test point (0,0)
0>-02+4(0)-3
0>-3
x y
0 -3
1 0
2 1
3 0
4 -3
Test Point
Last Example! Sketch the intersection of the given inequalities.1 y≥x2 and 2 y≤-x2+2x+4
Graph both on the same coordinate plane. The place where the shadings overlap is the solution.
Vertex of #1: (0,0)Other points: (-2,4), (-1,1),
(1,1), (2,4)
Vertex of #2: (1,5)Other points: (-1,1), (0,4), (2,4),
(3,1)
* Test point (1,0): doesn’t work in #1, works in #2.
SOLUTION!
Solving Quadratic Inequalities
Solve the quadratic inequality Solve the quadratic inequality xx2 2 – 5– 5x x + 6 > 0 graphically.+ 6 > 0 graphically.
Procedures:
Step (2): we have y = (x – 2)(x – 3) ,i.e. y = 0, when x = 2 or x = 3.
Factorize x2 – 5x + 6,
The corresponding quadratic function is y = x2 – 5x + 6
Sketch the graph of y = x2 – 5x + 6.
Step (1):
Step (3):
Step (4): Find the solution from the graph.
Sketch the graph Sketch the graph y =y = xx2 2 – 5– 5x x + 6 .+ 6 .
x
y
06 5
2 x x y
What is the solution of What is the solution of xx2 2 – 5– 5x x + 6 > + 6 > 0 0 ??
y = (x – 2)(x – 3) , y = 0, when x = 2 or x = 3.
2 3
above the x-axis.so we choose the portion
x
y
0
We need to solve x 2 – 5x + 6 > 0,
The portion of the graph above the x-axis represents y > 0 (i.e. x 2 – 5x + 6 > 0)
The portion of the graph below the x-axis represents y < 0 (i.e. x 2 – 5x + 6 < 0)
2 3
x
y
0
6 52
x x y
When x < 2x < 2,the curve is
above the x-axisi.e., y > 0
x2 – 5x + 6 > 0
When x > 3x > 3,the curve is
above the x-axisi.e., y > 0
x2 – 5x + 6 > 0
2 3
From the sketch, we obtain the solution
3xor2x
Graphical Solution:
0 2 3
Solve the quadratic inequality Solve the quadratic inequality xx2 2 – 5– 5xx + 6 < 0 graphically. + 6 < 0 graphically.
Same method as example 1 !!!Same method as example 1 !!!
x
y
0
6 52
x x yWhen 2 < x < 32 < x < 3,
the curve isbelow the x-axis
i.e., y < 0x2 – 5x + 6 < 0
2 3
From the sketch, we obtain the solution
2 < x < 3
0 2 3
Graphical Solution:
Solve
Exercise 1:
.012 xx
x < –2 or x > 1
Answer:
x
y
0
1 2 x x y
0–2 1
Find the x-intercepts of the Find the x-intercepts of the curve:curve:
(x + 2)(x – 1)=0(x + 2)(x – 1)=0
x = –2 or x = 1x = –2 or x = 1
–2 1
Solve
Exercise 2:
.0122 xx
–3 < x < 4
Answer:
x
y
0
122
x x y
0–3 4
Find the x-intercepts of the curve:Find the x-intercepts of the curve:
xx22 – x – 12 = 0 – x – 12 = 0
(x + 3)(x – 4)=0(x + 3)(x – 4)=0
x = –3 or x = 4x = –3 or x = 4
–3 4
Solve
Exercise 3:
.107
22
xx
–7 < x < 5
Solution:
x
y
0
35 22
x x y
0–7 5
Find the x-intercepts of the Find the x-intercepts of the curve:curve:
(x + 7)(x – 5)=0(x + 7)(x – 5)=0
x = –7 or x = 5x = –7 or x = 5
10
7
22
xx
271022 xx
03522 xx
057 xx–7 5
Solve
Exercise 4:
.3233 xxx
Solution:
x
y
0
35 22
x x y
Find the x-intercepts of the Find the x-intercepts of the curve:curve:
(x + 3)(3x – 2)=0(x + 3)(3x – 2)=0
x = –3 or x = 2/3x = –3 or x = 2/3
3233 xxx
03233 xxx
0233 xx
–3 23
0–3 23
x –3 or x 2/3