Problem 13

Question

Find a second-degree polynomial $p(x)=a x^{2}+b x+c$ with $p(0)=3, p^{\prime}(0)=2$, and $p^{\prime \prime}(0)=6 .$

Step-by-Step Solution

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Answer

The polynomial is $ p(x) = 3x^2 + 2x + 3 $.

1Step 1: Understand the Problem

We are asked to find the coefficients of a second-degree polynomial $ p(x) = ax^2 + bx + c $, given the value of $ p(0) $, the first derivative $ p'(0) $, and the second derivative $ p''(0) $.

2Step 1: Use the Given Condition for $ p(0) $

The condition $ p(0) = 3 $ implies that when we substitute $ x = 0 $ into $ p(x) = ax^2 + bx + c $, the result is 3. Therefore, $ c = 3 $.

3Step 2: Use the Given Condition for the First Derivative $ p'(0) $

First, find the derivative of $ p(x) $. The first derivative is $ p'(x) = 2ax + b $. The condition $ p'(0) = 2 $ implies that when we substitute $ x = 0 $ in $ p'(x) = 2ax + b $, the result is 2. So, $ b = 2 $.

4Step 3: Use the Given Condition for the Second Derivative $ p''(0) $

Now find the second derivative of $ p(x) $. The second derivative is constant: $ p''(x) = 2a $. The condition $ p''(0) = 6 $ implies $ 2a = 6 $, thus $ a = 3 $.

5Step 4: Write the Polynomial

Now that we have the values of $ a $, $ b $, and $ c $, substitute them back into the polynomial form: $ p(x) = 3x^2 + 2x + 3 $.

Key Concepts

Polynomial CoefficientsDerivative of a PolynomialSecond Derivative

Polynomial Coefficients

In algebra, a polynomial is an expression made up of variables (like $x$) and coefficients, along with operations like addition, subtraction, and multiplication. Each part of a polynomial that is separated by a plus or minus sign is called a term. A second-degree polynomial specifically has the highest variable raised to the power of two, and looks like this:

$p(x) = ax^2 + bx + c$

Here:

$a$ is the coefficient of the $x^2$ term
$b$ is the coefficient of the $x$ term
$c$ is the constant term or the polynomial value at $x=0$

The coefficients $a$, $b$, and $c$ determine the specific shape and position of the parabola that the polynomial represents. Finding these coefficients given certain conditions, as in this exercise, is crucial in understanding how specific changes in them affect the graph of the polynomial.

Derivative of a Polynomial

The derivative of a function measures how the function changes as its input changes. For a polynomial, taking a derivative means finding a new polynomial that gives the slope of the tangent line to the original polynomial at any point. For our polynomial:

$p(x) = ax^2 + bx + c$

The first derivative $p'(x)$ is found by applying the power rule to each term:

The derivative of $ax^2$ is $2ax$ because you bring down the exponent (2) and then reduce it by 1.
The derivative of $bx$ is $b$ since the derivative of any constant times a variable raised to the first power is the constant itself.
The derivative of a constant, $c$, is 0.

Putting it all together:

$p'(x) = 2ax + b$

This derivative function allows us to determine the rate of change at any given point on the original polynomial. For our exercise, knowing $p'(0) = 2$ helped us find the value of $b$.

Second Derivative

The second derivative of a function gives us the rate at which the first derivative, or the slope, is changing. In simple terms, it tells us how quickly the curve of the function is bending or the function's acceleration. For our polynomial's first derivative:

$p'(x) = 2ax + b$

To find the second derivative, $p''(x)$, we differentiate $p'(x)$ again:

The derivative of $2ax$ is $2a$, which is a constant. It comes from applying the power rule, where applying it to $2ax^1$ yields $2a \times x^{0} = 2a$.
The derivative of $b$, a constant, is simply 0.

Thus, the second derivative becomes:

$p''(x) = 2a$

In our example, knowing $p''(0) = 6$ allowed us to solve for $a$, reaffirming that the bend or concavity of the parabola is consistent with the original function's shape.

Problem 13

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