SWBAT: Graph the Systems of · PDF fileSWBAT: Graph the Systems of Equations A quadratic-linear system consists of a quadratic equation and a linear equation. The solution ... (7,5);
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SWBAT: Graph the Systems of Equations
Step 1: Calculate the Midpoint.
πππ = (ππ + π
π,
βπ + βπ
π)
πππ = (π, βπ)
Step 2: Calculate the radius of the circle.
ΞX ΞY
r2 = βπ π+ βπ π
10-4 -6
-9-(-5) -4
r2 = βπ π+ βπ π
r2 = ππ
(x - 7)2 + (y + 7)2 = 52
SWBAT: Graph the Systems of Equations
A quadratic-linear system consists of a quadratic equation and a linear equation. The solution of a quadratic-linear system is the set of ordered pairs of numbers that make both equations true. As shown below, the line may intersect the curve in two, one, or no points. Thus the solution set may contain two ordered pairs, one ordered pair, or no ordered pairs.
#1) Quadratic Linear System of Equations
SWBAT: Graph the Systems of Equations
Vertex = (2,-1)
2 -1
3
1
4
5 6
0
-1 -2
0
3
8 15
0
3
8
15
1
3
Rule: ππ
π,
ππ
π,
ππ
π,
ππ
π,
ππ
π
-4
1
π
π = π
βπ
Quadratic Linear System of Equations:
SWBAT: Graph the Systems of Equations
Rule: ππ
π,
ππ
π,
ππ
π,
ππ
π,
ππ
π
π
π,π
π,π
π,π
π,π
π
#2) Quadratic Linear System of Equations:
SWBAT: Graph the Systems of Equations
Vertex = ( , )
x + y = 4
Rule: ππ
π,
ππ
π,
ππ
π,
ππ
π,
ππ
π
#2) Quadratic Linear System of Equations:
SWBAT: Graph the Systems of Equations
1 5
2
0
3
4 5
-1
-2 -3
4
1
-4 -11
4
1
-4
-11
-1
4 Vertex = (1,5)
Rule: ππ
π,
ππ
π,
ππ
π,
ππ
π,
ππ
π
x + y = 4
Rule: βπ
π,
βπ
π,
βπ
π,
βπ
π,
βπ
π
#2) Quadratic Linear System of Equations
SWBAT: Graph the Systems of Equations
SWBAT: Solve a Non-Linear System of Equations
π₯2 + π¦2 = 4 π¦ = π₯
Step 1: Substitute βyβ with βxβ
π₯2 + (π₯)2= 4
Step 2: Solve for x
π₯2 + π₯2 = 4
2π₯2 = 4 2 2
π₯2 = 2
π₯2 = 2
π₯ = Β± 2
Step 3: Find y for each x.
π¦ = π₯ πβππ π₯ = 2
y = 2 πβππ π₯ = β 2
y = - 2
Solutions = ( 2, 2) and (β 2,β 2)
SWBAT: Graph the Systems of Equations
Solutions = ( 2, 2) and (β 2,β 2)
Solutions (π. π, 1.4) and (β1.4,β1.4)
SWBAT: Graph the Systems of Equations
πΉπππ π‘ ππππππ: (π₯ β 3)2+ (π¦ β 5)2= 25
c: (3,5); r = ππ = π
ππππππ ππππππ: (π₯ β 7)2+ (π¦ β 5)2= 9
c: (7,5); r = π = π
Solutions = (π, π) and (7,2)
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