# The second degree polynomial.

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 3.1, Problem 61E
To determine

## To find: The second degree polynomial.

Expert Solution

The second degree polynomial is P(x)=x2x+3_.

### Explanation of Solution

Given:

The second degree polynomial satisfies P(2)=5 ,P(2)=3 and P(2)=2.

Derivative rules:

(1) Constant multiple rule: ddx(cf)=cddx(f)

(2) Power rule: ddx(xn)=nxn1

(3) Sum rule: ddx(f+g)=ddx(f)+ddx(g)

Calculation:

The general form of the second degree polynomial is P(x)=ax2+bx+c (1)

Obtain the first derivative of P(x)=ax2+bx+c.

P(x)=ddx(P(x))=ddx(ax2+bx+c)

Apply the sum rule (3) and the constant multiple (1),

Since the derivative of constant function is zero, ddx(c)=0.

Apply the power rule (2) and simplify the terms,

P(x)=a(2x21)+b(1x11)=2a(x)+b(1)=2ax+b

Therefore, the first derivative of P(x) is P(x)=2ax+b_.

Obtain the second derivative of P(x)=ax2+bx+c.

P(x)=ddx(P(x))=ddx(2ax+b)

Apply the sum rule (3) and the constant multiple (1),

Since the derivative of constant function is zero, ddx(b)=0.

Apply the power rule (2) and simplify the terms,

P(x)=2a(1x11)=2a

Therefore, the second derivative of P(x) is P(x)=2a_.

Obtain the second degree polynomial P.

Substitute 2 for x in P(x)=2a,

P(2)=2a

Since P(2)=2, 2a=2.

Thus, the value of a=1.

Substitute 2 for x in P(x)=2ax+b,

P(2)=2a(2)+b=4a+b=4(1)+b     [Qa=1]=4+b

Since P(2)=3, 4+b=3.

b=34=1

Thus, the value of b=1 .

Substitute 2 for x in P(x)=ax2+bx+c,

P(2)=a(2)2+b(2)+c=4a+2b+c=4(1)+2(1)+c    [Qa=1,b=1]=2+c

Since P(2)=5, 2+c=5.

c=52=3

Thus, the value of c=2.

Substitute the values 1 for a,-1 for b and 3 for c in P(x)=ax2+bx+c,

P(x)=(1)x2+(1)x+3=x2x+3

Therefore, the second degree polynomial is P(x)=x2x+3_.

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