How to Use Properties of Polynomial Graphs Questions with Detailed Solutions

How to use the properties of the polynomial graphs to identify polynomials. Grade 12 math questions with detailed solutions and graphical interpretations are presented.

 Give four different reasons why the graph below cannot possibly be the graph of the polynomial function $p(x) = x^4-x^2+1$. . Solution The four reasons are: 1) The given polynomial function is even and therefore its graph must be symmetric with respect to the y axis. The given graph is not symmetric with respect to the y axis. 2) The given polynomial function does not have real zeros (discriminant = -3 : negative). The given graph has x intercepts which must corresponds to real zeros. 3) The y intercept calculated using p(x)( p(0) = 0^4-0^2+1 = 1) is positive. The y intercept of the graph is negative. 4) Having a leading coefficient ( = 1) positive and an even degree ( = 4), the polynomial must have a graph with the right and left both rising. In the given graph, they are both falling. Match the polynomial functions to their graphs where all x intercepts are shown. $f(x) = (x+1)(x-1)^2(x+2)^2$ $g(x) = -(x+1)(x-1)^4$ $h(x) = (x+1)(x-1)^3(x-3)$ $i(x) = (x+1)^2(x-2)^3$ $j(x) = (x+1)^2(1-x)(x-2)^2$ $k(x) =-(x+1)^2(x-1)^2(x-3)$ . Solution According to their equations, all 6 given polynomial functions are of degree 5. However their leading coefficients are of different signs. We classify the 6 polynomials into 2 groups: I and II Group I - Given polynomials with positive leading coefficients f(x) = (x + 1)(x - 1)2(x + 2)2 h(x) = (x + 1)(x - 1)3(x - 3) i(x) = (x + 1)2(x - 2)3 Having degree 5 (odd) and leading coefficients positive, each of the graphs of the above polynomials (f, h and i) has the following graphical properties: as   x ____> ∞   ,   y ____> ∞   (the right hand side of the graphs rises) as   x ____> - ∞   y ,   ____> - ∞   (the left hand side of the graph falls) The given graphs in parts a) c) and e) have the above properties with different x intercepts and their multiplicities. Hence Polynomial f(x) = (x + 1)(x - 1)2(x + 2)2 has a zero of multiplicity 1 at x = -1 , a zero of multiplicity 2 at x = 1 and a zero of multiplicity 2 at x = - 2 and should correspond to the graph in part e). Polynomial h(x) = (x + 1)(x - 1)3(x - 3) has a zero of multiplicity 1 at x = -1 , a zero of multiplicity 3 at x = 1 and a zero of multiplicity 1 at x = 3 and should correspond to the graph in part a). Polynomial i(x) = (x + 1)2(x - 2)3 has a zero of multiplicity 2 at x = -1 and a zero of multiplicity 3 at x = 2 and should correspond to the graph in part c). Group II - Given polynomials with negative leading coefficients The polynomial functions g, j and k, when expanded, have leading coefficients that are negative. g(x) = - (x + 1)(x - 1)4 j(x) = (x + 1)2(1 - x)(x - 2)2 k(x) = - (x + 1)2(x - 1)2(x - 3) Having degree 5 (odd) and leading coefficients negative, each of the graphs of the above polynomials (g, j and k) has the following graphical properties: as   x ____> ∞   ,   y ____> - ∞   (the right hand side of the graphs falls) as   x ____> - ∞   y ,   ____> ∞   (the left hand side of the graph rises) The given graphs in parts b) d) and f) have the above properties with different x intercepts and their multiplicities. Hence Polynomial g(x) = - (x + 1)(x - 1)4 has a zero of multiplicity 1 at x = -1 , a zero of multiplicity 4 at x = 1 and should correspond to the graph in part f). Polynomial j(x) = (x + 1)2(1 - x)(x - 2)2 has a zero of multiplicity 2 at x = -1 , a zero of multiplicity 1 at x = 1 and a zero of multiplicity 2 at x = 2 and should correspond to the graph in part d). Polynomial k(x) = - (x + 1)2(x - 1)2(x - 3) has a zero of multiplicity 2 at x = -1 , a zero of multiplicity 2 at x = 1 and a zero of multiplicity 1 at x = 3 and should correspond to the graph in part b).

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