Parabolic Investigation

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A Parabola is a symmetrical open plane curve formed by the intersection of a cone with a plane parallel to its side. The path of a projectile under the influence of gravity ideally follows a curve of this shape. This is u

In this task I will investigate the patterns in the intersections of parabolas and the lines y = x and y = 2x. Forming a conjecture that holds true for the vertex of the parabola being in the first quadrant and then change it so it holds true for the vertex is in any quadrant. Then I will prove my conjectures for other lines like y = 3x and 4x and so on and I will also change the degree of the polynomials and their values to prove the conjuncture to be true for values greater than 3.

Using the dynamic graphing software GeoGebra construct the required graph

Graphical solution for the given equation is given below

(Fig 1.1)

1. ‘Find the four Intersections made my the parabola x2-6x+11 and the lines y=x and y=2x’

The co-ordinates for the intersections of the parabola and the two given lines are

Ans 1. To find the co-ordinates using technology graph the parabola and the two lines required, and note the points of intersection.

Alternatively solve the quadratic equation but substituting the value of y = x and y = 2x

Giving the two equations

To find Equation 1 - (x2 – 7x + 11 = 0)

Change (y = x2 – 6x + 11) to (x = x2 – 6x + 11) by substituting x by y and solve the quadratic equation. The solution for the equation one will give the points (x2, y2) and (x3, y3)

To find Equation 2 - (x2 – 8x + 11 = 0)

Change (y = x2 – 6x + 11) to (2x = x2 – 6x + 11) by substituting 2x by y and solve the quadratic equation. The solution for the equation one will give the points (x1, y1...

... middle of paper ...

...19 0.13 -0.17 0.30 0

The value for D turns out to be zero for all values of the cubic polynomials thereby preventing us to form a conjecture for D = 0

Therefore we can not find a conjecture like the conjecture for a second degree polynomial

6. ‘Consider weather the conjecture might be modified to include higher order polynomials’

Ans 6.

Using the dynamic graphing software GeoGebra construct the required graph.

Below the graph consist of the intersections of the line ‘y = x’ , ‘y = 2x’ with the curve y = x4

The graph has 4 points of intersections with the ‘y=x’ and ‘y=2x’.

Like the graph of x2 the graph of x4 provides four points of intersection.

Therefore we can use the formulae D = | SL - SR |

Bibliography

IGCSE Mathematics – Pimentel & Wall

IBDP Press Mathematics Higher Level [Core] – Nigel Buckle & Iain Dunbar

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