Chapter Closure
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Exercise 173 Page 366
a To write the equation in graphing form we have to complete the square. In order to do this, the coefficient of x^2 has to be 1.
b The equation is describing a circle.
a Graphing Form: y=3(x-5)^2+(- 2)
Graph:
Domain: All real numbers.
Range: y≥ - 2
Is It a Function? Yes.
b Graphing Form: (x-3)^2+(y-(- 2))^2=9
Graph:
Domain: 0≤ x ≤ 6
Range: - 5≤ y ≤ 1
Is It a Function? No.
a A quadratic in graphing form is written on the following format.
y/3=x^2-10x+73/3
AddEqn
LHS+(- 10/2)^2=RHS+(- 10/2)^2
y/3+(- 10/2)^2=x^2-10x+73/3+(- 10/2)^2
▼
Simplify
NegBaseToPosBase
(- a)^2=a^2
y/3+(10/2)^2=x^2-10x+73/3+(10/2)^2
CommutativePropAdd
Commutative Property of Addition
y/3+(10/2)^2=x^2-10x+(10/2)^2+73/3
CalcQuot
Calculate quotient
y/3+5^2=x^2-10x+5^2+73/3
SplitIntoFactors
Split into factors
y/3+5^2=x^2-2(x)(5)+5^2+73/3
FacNegPerfectSquare
a^2-2ab+b^2=(a-b)^2
y/3+5^2=(x-5)^2+73/3
y=3(x-5)^2+(-2)
Is it a function?
Since there is no part of the graph where one x-value gives multiple y-values, we know that it is a function.
Domain and Range
There are no x-value that we cannot substitute into the function, which means the domain must be all real numbers. The range shows the set of y-values or outputs a function gives. Since the function's minimum value is y=- 2, the range must be y≥ - 2. Domain:& All real numbers Range:& y≥ - 2
b Examining the equation, we notice that both y and x are squared. This means that this equation is describing a circle. The general equation of a circle can be written in the following way.
x^2+y^2-6x+4y+4=0
AddEqn
LHS+(- 6/2)^2=RHS+(- 6/2)^2
x^2-6x+y^2+4y+4+(- 6/2)^2=(- 6/2)^2
AddEqn
LHS+(4/2)^2=RHS+(4/2)^2
x^2-6x+y^2+4y+4+(- 6/2)^2+(4/2)^2=(- 6/2)^2+(4/2)^2
▼
Simplify
CalcQuot
Calculate quotient
x^2-6x+y^2+4y+4+(- 3)^2+2^2=(- 3)^2+2^2
NegBaseToPosBase
(- a)^2=a^2
x^2-6x+y^2+4y+4+3^2+2^2=3^2+2^2
CommutativePropAdd
Commutative Property of Addition
x^2-6x+3^2+y^2+4y+2^2+4=3^2+2^2
SplitIntoFactors
Split into factors
x^2-2(x)(3)+3^2+y^2+2(y)(2)+2^2+4=3^2+2^2
FacNegPerfectSquare
a^2-2ab+b^2=(a-b)^2
(x-3)^2+y^2+2(y)(2)+2^2+4=3^2+2^2
FacPosPerfectSquare
a^2+2ab+b^2=(a+b)^2
(x-3)^2+(y+2)^2+4=3^2+2^2
CalcPow
Calculate power
(x-3)^2+(y+2)^2+4=9+4
AddTerms
Add terms
(x-3)^2+(y+2)^2+4=13
SubEqn
LHS-4=RHS-4
(x-3)^2+(y+2)^2=9
a+b=a-(- b)
(x-3)^2+(y-(- 2))^2=9
The definition of a function is a graph where one input corresponds to only one output. We can check if this is the case by performing a Vertical Line Test. If we draw a vertical line anywhere in the diagram, it should never hit the graph more than once if it's a function.
As we can see, the graph failed the Vertical Line Test, and therefore it is not a function. The domain tells us the x-values for which the graph is defined, and the range is the y-values for which the graph is defined. Examining the diagram, we can identify the range and domain.
With this information, we can write the domain and range. Domain:& 0≤ x ≤ 6 Range:& -5≤ y ≤ 1
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