Grade 12 Find a Polynomial Given its Graph Questions with Detailed Solutions

How to find a polynomial given its graph. Grade 12 math questions are presented along with their detailed solutions.

 Find the equation of the cubic polynomial function g shown below. . Solution 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)2, hence function g has the equation: g(x) = k (x + 1)(x - 3)2 , where k is a constant. Constant k may be found using the point with coordinates (1 , 3) shown in the graph. g(1) = k (1 + 1)(1 - 3)2 = 3 Simplify and solve for k. k = 3 / 8 g(x) is given by. g(x) = (3 / 8)(x + 1)(x - 3)2 Find the fourth-degree polynomial function f whose graph is shown in the figure below. . Solution 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)2, hence the polynomial f has the equation: f(x) = k (x - 2)(x + 2)(x + 1)2 , where k is a constant. 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)2 = -1 Simplify and solve for k. k = 1 / 4 f(x) is given by. f(x) = (1/4)(x - 2)(x + 2)(x + 1)2 Find the equation of the degree 4 polynomial f graphed below. . Solution 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)3 (x - 5 / 2) , where k is a constant. We now use the point (2 , -4) to fink k. - 4 = k(2)3 (2 - 5 / 2) , solve for k: k = 1 The equation of polynomial f(x) is given by. f(x) = x3 (x - 5 / 2) The graph of a cubic polynomial $y = a x^3 + b x^2 +c x + d$ is shown below. Find the coefficients a, b, c and d. . Solution 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)2. Hence the polynomial may be written as y = k(x + 2)(x - 1)2 We now need to find k using the y -intercept (0 , 1) shown in the graph. 1 = k(0 + 2)(0 - 1)2 = 2 k 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)2 = 0.5 x3 - 1.5 x + 1 We now compare the expression of the polynomial found above to y = a x3 + b x2 +c x + d and obtain the values of the coefficients a = 0.5 , b = 0 , c = -1.5 and d = 1 The graph of polynomial $y=ax^4+bx^3+cx^2+dx+e$ is shown below. Find the coefficients b, d and e. . Solution 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 x3   and   d x   included in the given expression of the polynomial above are not even and therefore their coefficients are equal to 0. Hence b = 0 , d = 0 and therefore y is given by y = a x4 + c x2 + e Coefficient e is found using the y intercept (0 , -2) from the graph. -2 = a (0)4 + c (0)2 + e e = -2

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