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Graphs of Polynomial Functions Project
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Graphs of Polynomial Functions Project
Finding the Equation for this Graph
Because the zeros 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
Finding the Equation for this Graph
Kevin Huang
Graph of a 3rd Degree Polynomial
This graph has a root of multiplicity 1
and a root of multiplicity 2
Because the zeros of this equation 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 the Equation for this Graph
Because the zeros of this equation are -3 (with multiplicity 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 of a 4th Degree Polynomial
This graph has 2 roots of multiplicity 1
and a root of multiplicity 2
Graph #2
Graph of a 2nd Degree Polynomial
Graph #1
This is a 5th degree graph with a root of multiplicity 3 and 2 roots of mulplicity 1
Finding the Equation for Graph #1
This graph has
2 roots of multiplicity 1
Finding the Equation for Graph #2
This is a 4th degree graph with a root of multiplicity 2 and 2 roots of mulplicity 1
Because the zeros 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 zeros 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)