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# Designer Polynomials

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## Brooke Tryon

on 24 May 2011

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#### Transcript of Designer Polynomials

Quadratic Functions of the form f(x)=x²^2+cx+d (1) How many critical points are
there for each choice of c and d? Good Morning Class!
Today we will investigate
the relationships between
polynomial functions and their
graphs. The specific
polynomials we will be studying
and quartic functions. Find derivative: 2x+c
Set equal to zero and solve for x:
2x+c=0
x=-c/2

For each choice of c (where d always equals zero), there will be only one value x that satisfies c = -2x. Therefore, there is only one critical point for each choice c and d. f(x)=x^2+cx+d

(2) Are the critical points maxima, minima, or neither? Since the function is quadratic, the graph is
a parabola. Because there is no coefficient
in front of the x-squared term, it cannot be
negative. Therefore, the function f(x) must
always be concave up, making the one
critical point described in question (1) a
minimum. Graph of x^2 Graph of -x^2 (3) How should you choose c and d to insure
that all critical points occur at integer values
of x? f'(x) = 2x + c
2x + c = 0
x = -c/2 For any value c, -c/2 must be an integer. For this to be
true, c must be any even number, positive or negative, so
that when you divide it by 2, you still get an integer. (4) What must the shape of the graph of f be? f''(x) = 2
Since the second derivative cannot be negative,
the function is always concave up. This, along
with the fact that f has only one critical point means
that the graph of f is a parabola. B. Cubic functions of the form
g(x) = x^3 + bx^2 + cx + d (5) How many inflection points are there? g'(x) = 3x^2 + 2bx + c

g''(x) = 6x + 2b =0
b = -3x b is the only coefficient that affects inflection points, so for
any value b, there is only one value x that satisfies b = -3x,
resulting in only one candidate for an inflection point. (6) Produce examples where the inflections points and critical points
all occur at integer values of x where g(x) has: (a) 0 critical points, (b)
1 critical point, and (c) 2 critical points. a. No Critical Points
Critical Point Test
g’(x) = 3x^2 + 2bx + c
Let b = 3 and c = 8
g’(x) = 3x^2 + 6x + 8 = 0
Discriminant: x = (-60)1/2
No real zeros. No critical points exist.

Inflection Point Test
g”(x)=6x+6=0
x= -1
One inflection point: x= -1 b. One Critical Point
Critical Point Test
g’(x)=3x2+2bx+c
Let b=0 and c=0
g’(x)=3x2=0
x=0
One critical point exists: x=0

Inflection Point Test
g”(x)=6x=0,
x=0
One inflection point: x=0 c. Two Critical Points
Critical Point Test
g’(x)=3x2+2bx+c
Let b= -3 and c= -9
g’(x)=3x2-6x-9=0, (3x+3)(x-3)=0
x= -1 and x= 3
2 critical points: x= -1 and x= 3

Inflection Point Test
g”(x)=6x-6=0
x= 1
One inflection point: x= 1 (7) Write general rules for choosing b, c, and d to produce families of
examples in question 6. Use the discriminant! (b^2-4ac)^1/2
Then modify the discriminant using the derivative 3x^2 + 2bx + c to determine the number of critical points:
(b^2-4ac)^1/2 turns into ((2b)2-12c)1/2.
You can ignore the square root sign and apply the power to make 4b^2-12c, then 4b^2=12c, then finally b^2=3c. If b^2 = 3c, the function will have no real critical points. If b^2 < 3c, function will have exactly one real critical point with an integer value. If b^2 >3c, and if -c = b^2, the function will have two real critical points with integer values. C. Quartic functions of the form h(x) = x^4 + ax^3 + bx^2 + cx + d (8a) Choose a, b, c, and d so that h(x)
has 3 critical points at integer values
of x. let a = 8, b = 6, c = –40, and d = 12.
Verification: h(x) = x^4 + 8x^3 + 6x^2 – 40x + 12 and
h'(x) = 4x^3 + 24x^2 + 12x – 40. The derivative factors into the form (2x + 4)(2x – 2)(x + 5). Setting this equal to zero, we can solve for x:
x = –2, 1, –5
These are all integer values! (8b) Produce some general techniques
to generate a family of examples with 3
critical points, all at integer values of x. We know that the derivative is h'(x) = 4x^3 + 3ax^2 + 2bx + c. However, the best approach is not to start here, but to begin with the roots and work back to the derivative.

For example, start with (2x – 4)(2x + 2)(x – 1), being careful to make sure that each can be solved to get an integer value of x. This foils into 4x^3 – 8x^2 – 4x + 8. Solving for the critical points, we get x = 2, –1, 1. All integers! Assumptions For this project to be possible, some assumptions
had to be made. These include that students already have basic calculus knowledge, like how to differentiate and solve functions to find their zeros, as well as have knowledge of fundamental pre-calculus standards, like how to use the discriminant of a function. Predictions The major prediction of this project is that the students will learn the polynomials content so well that they will excel in all polynomial-related things for the rest of their lives. If we, as their teachers, were to give them a test over this material, they would all make A’s! Conclusions >The critical points of a function are found through the derivative of the function, and the inflection points are found through the second derivative of that function. Functions can have different numbers of critical points depending on their characteristics.
>Functions have different graphical shapes, depending on the power of the function.
>A course of action would be to assign a project that assesses the students’ comprehension of the material.
>A Foreseeable implications of this action plan is that the students will apply this knowledge to their everyday lives, in solving problems that involve functional representation. Thank you for your time, class!
It has been a pleasure being your
calculus teachers this beautiful morning.
This concludes our lesson on the
relationships between polynomial functions
and their graphs.
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