Helppp!!!13

Question

Sketch a possible graph for each set of characteristic used to define a particular polynomial function.

a) Degree of 2, one turning point which is a maximum, and a constant term of 3.

b) Degree of 2, one turning point which is a maximum, and only one x-intercept.

c) Two turning point, a positive leading coefficient,one x intercept and a constant term of 3.

d) A cubic function with two x intercepts, a negative leading coefficient and a constant term -2.

Answer 1

(a)

Consider the polynomial of degree 2 .

f(x) = ax² +bx +c

Given constant term c = 3 

f(x) = ax² +bx + 3

Given one turning point which is a maximum .

Now find the turning point , by equating the first derivative to zero .

f '(x) = 2ax +b

2ax +b = 0

2ax = -b 

x = -b/2a 

So the polynomial has a turning point at x = -b/2a .

f ''(x) = 2a

Since the given polynomial has turning point which is a maximum 2a < 0 .

So a<0 .

f(x) = -ax² +bx + 3

Now consider a polynomial with the following constraints .

f(x) = -2x² +4x + 3

a = -2 and b = 4

So the turning point is exist at x = -4/2(-2) = 1 .

Now graph the function .

Answer 2

(b)

Consider the polynomial of degree 2 .

f(x) = ax² +bx +c

Given only one x-intercept 

Let the x-intercept  is k .

So the function f(x) = a(x-k)² 

f(x) = a [x² -2kx +k² ]

f(x) = ax² -2akx +ak²

Given one turning point which is a maximum .

Now find the turning point , by equating the first derivative to zero .

f '(x) = 2ax +2ak

2ax +2ak = 0

2ax = -2ak 

x = -2ak/2a 

x = -k

So the polynomial has a turning point at x = -k.

f ''(x) = 2a

Since the given polynomial has turning point which is a maximum 2a < 0 .

So a<0 .

f(x) = -a(x-k)² 

Now consider a polynomial with the following constraints .

f(x) =  -2(x-3)² 

a = -2 and k = 3

So the turning point is exist at x =3 .

Now graph the function .

Answer 3

(d)

Consider a cubic function 

f(x) = ax³ +bx² +cx +d

Negative leading coefficient , so a < 0 .

Given constant term -2 .

So f(x) = -ax³ +bx² +cx -2

So now consider a function with two x intercepts

f(x) = -1(x+2)(x+1)² 

f(x) = -1(x+2)(x²+1+2x)

f(x) = -1(x³+x+2x²+2x²+2+4x)

f(x) = -1(x³+4x²+5x+2)

f(x) = -x³-4x²-5x-2

Now graph the function 

Answer 4

(c)

Graph the function with two turning point, a positive leading coefficient,one x intercept and a constant term of 3.

We know that if there are n turning points then the degree of the polynomial is (n+1) or more.

There are two turning points then the degree of the Equation can be 3 or greater.

Consider a Cubic Function

Now the function has one x-intercept,

then third degree equation is (x - a)(bx² + cx + d) = 0

Solving the Equation

bx³ + cx²+dx - abx² - acx - ad = 0

bx³ + (c - ab) x² + (d-ac) x - ad = 0

The cubic Function is bx³ + (c - ab) x² + (d-ac) x - ad = 0.

The Constant term is 3.

From the constant term from the cubic Equation is -ad

-ad = 3

ad = -3

Let us assume a = -3 and d = 1

The polynomial has positive leading coefficient which means b > 0

So Let us assume b = 1 and c = 2

Then Cubic Equation is

bx³ + (c - ab) x² + (d-ac) x - ad = 0

x³ + 5x² + 7x + 3 = 0

So the Cubic Equation is x³ + 5x² + 7x + 3 = 0

Graph

Therefore we can from the turning points are (-1,0) and (-2.33, -1.185) and x - intercept is (-3,0).