Math 111 Group Activity-Week 3B Polynomial Functions 1 Name: -Find the roots of
ID: 2903910 • Letter: M
Question
Math 111 Group Activity-Week 3B Polynomial Functions 1 Name: -Find the roots of p in factored form degree e determining domain/range interpreting factored form Sketch graphs of Identify multiplicity of polynomial roots in factored form and in graphical form factor e end behavier On a hike in the woods you stumble upon a patch of beautiful blackberries. You very much want to pick some to take home, but have no container with which to transport the berries. You do have a piece of 8.5 inch x 11 inch cardstock, a pair of scissors and tape. In a moment of brilliance, you realize that if you cut a square from each corner of the cardstock, it can be folded and taped into an open box (see the diagram at right). Wanting to take home as much as possible, your goal is to cut out the square so your box will have the maximum possible volume. Let the square cut from each corner have side length x (measured in inches). 1. Imagine cutting squares of varying size from each corner. Measure the side lengths and calculate the volume of the box, V. Record your data in the table below. a. Length of Cut 4.25 finches) Volume b. State the function that will model the volume of the box in terms of Vx) and its domain. -Explanation / Answer
1.a.
* It is not possible to cut corners of 6 inches side from a piece admeasuring 11 inches by 8.5 inches
b. Volume V (x) = length*breadth*height. Let x be the height ( i.e. the corners cut are of side x) Then V (x) = (11 -2x)( 8.5 -2x)x or, V (x) = x (11 -2x)(8.5 -2x) = 4x( x - 5.5) (x - 4.25)
2 a. V (x) = x [ 4x2 - 2x ( 11 + 8.5) + (11*8.5 )] = x ( 4x2 - 39x + 93.5) = 4x3 - 39x2 + 93.5x
b. The degree of V(x) is 3
c. The leading coefficient of V(x) is 4
3. a. V (x) is increasing in the intervals ( 0, 2) and ( 5.5, )
b. V (x) is dectreasing in the intervals ( 0, -) and (2, 5)
c. The local maxima of V 9x) is at x = 2 and is equal to 63
d. The local minima of V (x) is at x = 5 and is equal to - 7.5
4 a. V (x) will be 0 when x = 0, x = 4.25 and x = 5.5
b. The x - intercepts are the points where the graph crosses the x -axis, ie. where y = o i.e. the zeros of the function.
c. x = 5.5 does not make any sense as the breadth of the cardboard piece is only 8.5 inches. We can not cut corners of side 5.5 inches from this piece.
d. The largest possible value for V(x) is when x = 2. Then the volume of the box is 63 cubic inches.
5. We know that when the degree of a polynomial is even and
i. the graph rises both to the right and the left, the leading coefficient is positive.
ii. the graph falls both to the right and the left, the leading coefficient is negative
When the degree of a polynomial is even and
i. the graph rises to the right and falls to the left, the leading coefficient is positive
ii. the graph falls to the right and rises to the left, the leading coefficient is negative
a. The function is of odd defree and the leading coefficient is negative
b.The function is of even defree and the leading coefficient is positive
c.The function is of odd defree and the leading coefficient is positive
d.The function is of even defree and the leading coefficient is negative
6. f ( x) = a ( x+1)2 (x - 1) - graph A - odd degree and a < 0
k (x) = ax2 (x + 1) (x -1)- graph D a < 0
h (x) = ax(x+1)2(x -1)2 - graph C and a > 0
g (x) = a (x + 1)2(x -1)2 - graph- B and a > 0
7. The function has 1 zero at x = 2, 2 zeros at x = -2 and 2 zeros at x = -4. Its equation is (say) f (x) = a (x-2) (x+2)2(x + 4)2. Since its graph passes through the point ( -3, 2), we have 2 = a ( -5)(- 1)2(1)2 or, 2 = -5a. Therefore, a = -2/5. Thus the equation of the given graph is f (x) = -2/5 (x-2) (x+2)2(x + 4)2
Length of cut(inches) 1 2 4.25 6 Volume (cu. inches) 58.5 63 0 Not feasible*