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How to find a polynomial given its graph? Questions are presented along with their detailed solutions and explanations.

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The graph of the function has one zero of multiplicity 1 at x = -1 which corresponds to the factor x + 1 and and a zero of multiplicity 2 at x = 3 (graph touches but do not cut the x axis) which corresponds to the factor (x - 3)

g(x) = k (x + 1)(x - 3)

Constant k may be found using the point with coordinates (1 , 3) shown in the graph.

g(1) = k (1 + 1)(1 - 3)

Simplify and solve for k.

k = 3 / 8

g(x) is given by.

g(x) = (3 / 8)(x + 1)(x - 3)

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The graph of the polynomial has a zero of multiplicity 1 at x = 2 which corresponds to the factor (x - 2), another zero of multiplicity 1 at x = -2 which corresponds to the factor (x + 2), and a zero of multiplicity 2 at x = -1 (graph touches but do not cut the x axis) which corresponds to the factor (x + 1)

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

Constant k may be found using the y intercept f(0) = - 1 shown in the graph.

f(0) = k(0 - 2)(0 + 2)(0 + 1)

Simplify and solve for k.

k = 1 / 4

f(x) is given by.

f(x) = (1/4)(x - 2)(x + 2)(x + 1)

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The graph has x intercepts at x = 0 and x = 5 / 2. These x intercepts are the zeros of polynomial f(x). Because the graph crosses the x axis at x = 0 and x = 5 / 2, both zero have an odd multiplicity. The graph at x = 0 has an 'cubic' shape and therefore the zero at x = 0 has multiplicity of 3. The shape of the graph at x = 1/2 is close to linear hence the zero at x = 5 / 2 has multiplicity equal to 1. Using the zeros at x = 0 and x = 5 / 2, f(x) may be written as

f(x) = k (x - 0)

We now use the point (2 , -4) to fink k.

- 4 = k(2)

The equation of polynomial f(x) is given by.

f(x) = x

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The polynomial has degree 3. The graph of the polynomial has a zero of multiplicity 1 at x = -2 which corresponds to the factor x + 2 and a zero of multiplicity 2 at x = 1 which corresponds to the factor (x - 1)

y = k(x + 2)(x - 1)

We now need to find k using the y -intercept (0 , 1) shown in the graph.

1 = k(0 + 2)(0 - 1)

Solve for k.

k = 1 / 2

We now expand the polynomial, write it in standard form and identify the coefficient a, b, c and d.

y = (1 / 2)(x + 2)(x - 1)

We now compare the expression of the polynomial found above to

y = a x

and obtain the values of the coefficients

a = 0.5 , b = 0 , c = -1.5 and d = 1

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The graph of the polynomial is symmetric with respect to the y axis and therefore the polynomial function given above must be an even function. The terms b x

b = 0 , d = 0

and therefore y is given by

y = a x

Coefficient e is found using the y intercept (0 , -2) from the graph.

-2 = a (0)

e = -2

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