Assignment # 3

Further Explorations with Quadratics

by

Michael Ferra

Proposed InvestigationLet's reconsider the equation y = Ax

^{2}+ Bx + C.

i.Fix A = 1, C = 1, and y = 0. Observe different values of B in the xB plane to determine the roots.

ii.Fix A = 1, C = -1, and y = 0. Observe different values of B in the xB plane to determine the roots.

i. Fix A = 1, C = 1, and y = 0. Observe different values of B in the xB plane to determine the roots.Given A = 1, C = 1, and y = 0 for y = Ax

^{2}+ Bx + C, then 0 = x^{2}+ Bx + 1. Let's graph this in the xB plane.By taking any particular value of B and overlaying this equation on the graph we add a line parallel to the x-axis. If it intersects the curve in the xB plane, the intersection points correspond to the roots of the original equation for that value of B. Let's do this for a few values of B.

Observe from the graph, when B > 2 or when B < -2 that there are two roots to the original equation because the line intersects the curve twice. When B > 2, we get two negative real roots and when B < -2, we get two positive real roots. When B = 2, we get one real negative root and when B = -2, we get one positive real root. When -2 < B < 2, observe there are no real roots since there are no points of intersection. Is there an alternate way to find these roots? From our equation 0 = x

^{2}+ Bx + 1, the roots can be found by:Take the time to explore the different values of B in this equation to see that the above statements for B are in fact true! We can notice that when the discriminate is undefined, there are no real roots. Also notice that when the discriminate equals zero, that there is only one real root.

ii. Fix A = 1, C = -1, and y = 0. Observe different values of B in the xB plane to determine the roots.Given A = 1, C = -1, and y = 0 for y = Ax

^{2}+ Bx + C, then 0 = x^{2}+ Bx - 1. Let's graph this in the xB plane.By taking any particular value of B and overlaying this equation on the graph we add a line parallel to the x-axis. If it intersects the curve in the xB plane, the intersection points correspond to the roots of the original equation for that value of B. Let's do this for a few values of B.

Observe from the graph that at every value of B there are two intersections to the original equation because the horizontal line intersects the curve twice. Is there an alternate way to find these roots? From our equation 0 = x

^{2}+ Bx - 1, the roots can be found by:Take the time to explore the different values of B in this equation to see that the above statement for B is in fact true! We can notice that when the discriminate will never be undefined and will never equal zero, thus there will always be two real roots.