IB Mathematics Portfolio Type I

Candidate name: Ritesh Kothari

Candidate Code: 001859-018

School name: Mahatma Gandhi International School

School Code: 001859

- Graph the function f(x) = 4 - 8 3 + 182 - 12 + 24

The above graph is made using “Microsoft Excel 2007” and the displayed function is

F(x) = 4 – 83 + 182 – 12 + 24

Scale used: X: – 1.5 to 5.5

Y: 6 to 66

- Find the coordinates of the points of inflection Q and R. Determine the points P and S, where the line QR intersects the quartic function again, and calculate the ratio PQ:QR:RS.

Planning:

- Find the points of inflection in the function by differentiating the function twice.
- Substituting the X values into the function to get Y co-ordinates.
- Get the equation of a line which passes through the inflection points, named Q and R respectively.
- Find out two points P and S, where f() intersects the line which passes through the inflection points (Q and R).
- Finding out the roots of the function, ignoring the found points of inflection and use these roots to find other intersection points for the function.
- To find the ratios we will need to find out the distances between the points using the properties of similar triangles.
- Find the ratio between the points PQ, QR, and RS. Then simplify the points to get some result.

We find out the points of inflection of the given function, where the concavity of the graph changes. When the concavity of the function changes, then at that point the second derivative of the function is equal to zero.

Differentiating the given function:

F() = 4 – 83 + 182 – 12 + 24

F’() = 43 – 242 + 36 -12

Differentiate the function second time:

F’’(x) = 122 – 48 + 36

Solve the second derivative to get the values of x by equating it to zero.

=> 122 – 48 + 36

=> 2 – 4 + 3 = 0

=> ( – 3) ( – 1) = 0

=> = 3 and = 1

Concavity of the function changes at = 3 and = 1. To find out the co-ordinate points of these inflection points, substitute x values in the given function and find Y-co-ordinates.

Point Q: -

F() = 4 – 83 + 182 – 12 + 24

F(1) = 14 – 8(1)3 + 18(1)2 – 12(1) + 24 = 23

Point R:

F() = 4 – 83 + 182 – 12 + 24

F(3) = 34 – 8(3)3 + 18(3)2 – 12(3) + 24 = 15

Therefore, the two points of inflection are:

Q (1, 23)

R (3, 15)

To find the equation of a line which passes through the points Q and R, we first name the function as G(x). Equation of a line which passes from any two points can be found by the formula.

We will now substitute the known values (co-ordinate points) and find out the equation of line.

Y1 = 15

Y2 = 23

X1 = 3

X2 = 1

Therefore, the equation is:

Y = -4x + 27, where -4 is the slope of the line

=> G () = -4 + 27

Now our next step would be to find out the position of the points P and S. Points P and S are the intersection points of the functions f() and g(). By equating them we will get the X co-ordinates of the points on which f () and g () intersect.