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Solving Quadratic Equations with Graphs
Slideshow 31, MathematicsMr. Richard Sasaki, Room 307
Objectives
β’ Recall how to draw graphs with linear equations
β’ Be able to solve simultaneous equations graphically with both linear and quadratic lines
β’ Be able to do this with real world examples
Linear Graphs
First letβs have a review on how to draw linear graphs.
Linear graphs are in the form .π¦=ππ₯+πThey are straight lines and much easier to draw than quadratic graphs.ExampleDraw the graph .
Answers
Solving Simultaneous Equations Graphically
If we have a pair of simultaneous equations, do you remember how to solve them graphically?ExampleSolve the simultaneous equations and with the use of a graph.First draw both of the lines.The solution is simply the intersection. (2 ,3)So we get .π₯=2 , π¦=3
Answers
π₯=β2 , π¦=β4π₯=3 , π¦=β3π₯=2 , π¦=7
π₯=β1 , π¦=2π₯=β3 , π¦=β8π₯=2 , π¦=7
Whereβs the vertex?(0 ,β2)
Quadratic Simultaneous Equations
Letβs solve some simultaneous equations with quadratics like in Grade 8, but with graphs.ExampleSolve the simultaneous equations and graphically below:
11
The parabola allows for up to two intersections:
π₯=β1 , π¦=β1π₯=3 , π¦=7
Thatβs our two pairs of solutions!
Answers - Easy
π₯=β1 , π¦=β3π₯=3 , π¦=5
π₯=β3 , π¦=1π₯=0 , π¦=4
π₯=0 , π¦=β2π₯=5 , π¦=3
π₯=β3 , π¦=0π₯=2 , π¦=5
Answers - Medium
π₯=1 , π¦=2π₯=4 , π¦=8
π₯=β2 , π¦=β2π₯=2 , π¦=6
π₯=β2 , π¦=2π₯=β5 , π¦=8
π₯=0 , π¦=5π₯=2 , π¦=5
Answers - Hard
π₯=β4 , π¦=β3π₯=β1 , π¦=3
π₯=β9 , π¦=β7π₯=β6 , π¦=β4
π₯=β6 , π¦=0π₯=β2 , π¦=8
π₯=0 , π¦=β6π₯=4 , π¦=2
Real World Equations and Graphs Now we will look at some non-simultaneous real world problems.ExampleThe driver of a car travelling downhill on a road applies the brakes. The speed of the car (in ) is equal to where is the number of seconds after starting to brake.First, we need to draw the graph . (We donβt need negative values.)
π¦=β4 π₯2+12π₯+80π¦β80=β4(π₯ΒΏΒΏ2β3π₯+
94β94)ΒΏ
π¦=β4 (π₯β 32 )2
+89Vertex:
Real World Equations and Graphs Now we need to place that vertex on the grid and draw the line. Vertex: 41112
85
73
120
53
1
2825
How fast was the car travelling when the brakes were applied?80ππhβ 1
After how many seconds did the car reach its maximum speed? What was it?1.5 π ππππππ 89ππhβ 1
Answers - Easy
Answers - Hard πππππ ππ‘ππ¦ hπ‘ ππππ‘ππ£ππ‘π!
Graphing Real World Simultaneous EquationsUsually problems with simultaneous equations discuss their intersection. This could imply when a person catches up with another, an object lands on a slope or another meaning.ExampleA boy on a bicycle cycles down a slope and leaves the same time as his dog. The dog runs at and the boy on his bicycle has travelled metres. Draw a graph representing this and state the point in time when the boy passes the dog.
We should write the dogsβ statement in terms of how far it has travelled.
π¦=4 π₯
Graphing Real World Simultaneous EquationsNow we have both statements in terms of distance, and (where is the number of metres). Letβs use the formulae this time to find the vertex for .
h=β π2π,π=4 ππβπ
2
4π
h=ΒΏβ π2π
=ΒΏβ
2
(2β 12 )=ΒΏΒΏ
β2
π= 4ππβπ2
4 πΒΏ 0β2
2
4 β12ΒΏβ2
Vertex:
Graphing Real World Simultaneous EquationsLetβs draw the lines.
π¦=4 π₯π¦= π₯2
2+2π₯
ππππ‘ππ₯ :(β2 ,β2)Oh no, the vertex is off the graph! But itβs okay, because we know , so the next point isβ¦
222
62
10
(0 ,0)We know the boy catches up with his dog at the intersection, so it takesβ¦
4 π ππππππ
Answers - Easy
Answers - Hard