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Algebra 2 (Section 4.9)
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Algebra 2 (Section 4.9)
Review….The DISCRIMINANT is ________.
If b2 – 4ac > 0, then there are_______ REAL solutions and _______ x-intercepts.
acb 42
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REVIEW FROM 4.8 NOTES
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Algebra 2 (Section 4.9)
If b2 – 4ac < 0, then there are _______ REAL solutions and _______ x-intercepts.
NONO
REVIEW FROM 4.8 NOTES
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Algebra 2 (Section 4.9)
If b2 – 4ac = 0, then there IS _______ REAL solution and _______ x-intercept.
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REVIEW FROM 4.8 NOTES
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Algebra 2 (Section 4.9)
When there are __________ solutions, thereare _______ _______________________ .
NO REAL2 COMPLEX SOLUTIONS
REVIEW FROM 4.8 NOTES
Algebra 2 (Section 4.9)
Graph the following inequalities. Pick a test point, test it, and shade. (Show your work for testing!)
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Algebra 2 (Section 4.9)
STEPS to graphing a quadratic inequality Graph the parabola. In case you’ve forgotten how to do this…1. Find the vertex of the parabola
using .2. Substitute the x-coordinate into the
function to find the y-coordinate.
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a
bx
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Algebra 2 (Section 4.9)
3. Set up a table.4. Plot the points. *** THE NEW STUFF IS THE SHADING. ***5. Choose a point to test. (Pick one with
coordinates that are 0 if possible.)6. Shade.
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Algebra 2 (Section 4.9)
EX 1 Graph y < 2x2 + 8x – 1 .
9
228
ab
x2
2
128222 y
9y 9,2 V
92 71
34
1 70 1
Algebra 2 (Section 4.9)
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TEST (0,8)y < 2x2 + 8x – 18 < 2(0)2 + 8(0) – 18 < 0 + 0 – 1 8 < – 1 Shade the other side of the parabola!
FALSE!
Algebra 2 (Section 4.9)
EX 2 Graph
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1,4V14211
32
5 26 11
y x 3 4 12b g
This is in vertex form, so we don’t have to do work to get the vertex.
This parabola opensdown.
TEST (4,5)
5 > –3(4 – 4)2 + 15 > –3(0) + 15 > 0 + 15 > 1Shade that side of the parabola!
Algebra 2 (Section 4.9)
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y x 3 4 12b g
TRUE!
Algebra 2 (Section 4.9)
ASSIGNMENT #49WS
(Graphing Quadratic Inequalities)
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Algebra 2 (Section 4.9)
Systems of Quadratic Equations
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Algebra 2 (Section 4.9)
Give examples of what a system of equations involving quadratics (1 or
more) might look like graphically.
It could be ___ _______________ and a ________ .
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Algebra 2 (Section 4.9)
Give examples of what a system of equations involving quadratics (1 or
more) might look like graphically.It could be _____________.
It could be ___ _______________ and a ________ .
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1 point of intersection
Nointersection
2 points of intersection
2 QUADRATICS
Algebra 2 (Section 4.9)
172
1 point of intersection
Nointersection
2 points of intersection
It could be ____________ and______ .1 QUADRATIC 1 LINE
Algebra 2 (Section 4.9)Solve the system of equations.
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y x x
y x
RST2 12
2 8 Substitute!!!!!
2 8 122x x x x x2 2 x x 12 12
x x2 20 0 You can solve this by factoring or using the quadratic formula.
EX 1
Algebra 2 (Section 4.9)
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x x2 20 0 x x 5 4 0b gb g
x 5 0 x 4 0x 5 x 4
We have found the x values. Now we have to
find the y values.
We’re not done!
Algebra 2 (Section 4.9)
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y x x
y x
RST2 12
2 8
x 5 x 4y x 2 8y 2 5 8bg y 2 4 8bgy 18 y 0
5 18,b g 4 0,bg
Algebra 2 (Section 4.9)
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32
322
xy
xxy
32232 xxx
0 42 x x
2 2x x 3 3
0 4 x xb gx x 0 4
We’re not done!
EX 2
Algebra 2 (Section 4.9)
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32
322
xy
xxy
x x 0 4y x 2 3
y 2 4 3bgy 2 0 3bgy 3 y 50 3,b g 4 5,bg
Algebra 2 (Section 4.9)
23
23
2322
xy
xxy
232223 xxx 2 22 2x x 2 2 3 3x x
022 x022 x
02 x
02 x 0x
EX 3
Algebra 2 (Section 4.9)
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23
2322
xy
xxy
23 xy
0x 203 y2y
0 2,b g
Algebra 2 (Section 4.9)
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23
6522
xy
xxy
652223 xxx 3 3x x 2 2
88220 xx)442(20 xx
2220 xx2x
EX 4
Algebra 2 (Section 4.9)
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23
6522
xy
xxy
2x23 xy
223 y4y
2 4,b g
Algebra 2 (Section 4.9)
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ASSIGNMENT (Day 1)p.262(#8-12)
Your book says to solve by graphing. Do not do that! Solve algebraically
(like we did in this Power Point).
Algebra 2 (Section 4.9)
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432
22
xxy
xxyEX 5
06422 xx x x2 2 x x 2 2
22432 xxxx
0)322(2 xx 0132 xx
3x 1x
We’re not done!
Algebra 2 (Section 4.9)
293 4, b g
3x 1x
432
22
xxy
xxy
432 xxy
41321 y 43323 y4y 0y
1 0,b g
Algebra 2 (Section 4.9)
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ASSIGNMENT (Day 2) p.262(#14 – 18, 20,21)
