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Graphs of Polynomial Functions Project

Finding the Equation for Graph 1

Jonathan Gonzalez

Graph of a 3rd Degree Polynomial

Finding the Equation for this Graph

This graph has a root of multiplicity 1 and a root of multiplicity 2

Because the zeroes of this graph are -2 and 1, the equation must be

f(x)=a(x+2)(x-1)

To find a, we use the point in the graph (2,8)

8=a(2+2)(2-1)

8=a(4)(1)

8=4a

a=2

f(x)=2(x+2)(x-1)

Finding Equation for this Graph

Because the zeroes of this equation are -1, 1(with multiplicity 2), and 5 the equation must be

f(x)=a(x+1)(x-1)^(2)(x-5)

To find a we use the point in the graph (0,-10)

-10=a(0+1)(0-1)^(2)(0-5)

-10=a(1)(-1)^(2)(-5)

-10=-5a

a=2

f(x)=2(x+1)(x-1)^(2)(x-5)

Because the zeroes of this equation are -3 (with multiplicity of 2) and 2, the equation must be

f(x)=a(x+3)^2(x-2)

To find a, we use the point in the graph (0,-9)

-9=a(0+3)^2(0-2)

-9=a(9)(-2)

-9=-18a

a=.5

f(x)=.5(x+3)^2(x-2)

Graph 1

This is a 4th degree graph with a root of multiplicity2 and 2 roots of multiplicity 1

Graph of a 4th Degree Polynomial

Finding the Equation for Graph 2

This graph has 2 roots of multiplicity 1 and a root of multiplicity 2

Finding the Equation for this Graph

Because the zeroes of this equation are -2,0 (with multiplicity 3), and 2, the equation must be

f(x)+a(x+2)(x)^3(x-2)

To find a, we use the point in the graph (1,9)

9=a(1-2)(1-0)^3(1+2)

-9=a(-1)(1)^3(3)

-9=3a

a=-3

f(x)=-3(x+2)(x)^3(x-2)

Because the zeroes of this equation are -3, -2 and 1 (with multiplicity 2) the equation must be

f(x)=a(x+2)(x+3)(x-1)^2

To find a, we use the point in the graph (0,-6)

-6=a(0+2)(0+3(0-1)^2

-6=a(2)(3)(-1)^2

-6=6a

a=-1

f(x)=-(x+2)(x+3)(x-1)^2

Graph 2

This is a 5th degree graph with a root of multiplicity 3 and 2 roots of multiplicity 1

Graph of 2nd Degree Polynomial

This graph has 2 roots of multiplicity 1

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