Problem 32 - The Calculus with Analytic Geometry

Question

Find \(a, b\), and \(c\) so that the function defined by \(f(x)=a x^{2}+b x+c\) will have a relative maximum value of 7 at 1 and the graph of \(y=f(x)\) will go through the point \((2,-2)\).

Step-by-Step Solution

Answer

The values are a =-9, b =18, c = -2.

Step 1 - Identify the conditions

Given the function is of the form given by the equation: f(x)=a x^{2}+b x+c The conditions provided are: 1. The function will have a relative maximum value of 7 at x=1. 2. The graph of the function passes through the point (2,-2).

Step 2 - Use the relative maximum condition

Since the function has a relative maximum at x=1, we have: f(1) = 7 because the value of the function must be 7 at x=1. Substitute x=1 into the function: a(1)^2 + b(1) + c = 7 This simplifies to: a + b + c = 7 This is our first equation.

Step 3 - Use the derivative condition

For a relative maximum at x=1, the first derivative f'(x) must be zero at x=1. Find the first derivative of the function: f'(x) = 2ax + b Substitute x=1 into the derivative: f'(1) = 2a(1) + b Since f'(1) must be zero: 2a + b = 0 This is our second equation.

Step 4 - Use the point condition

Since the function passes through the point (2,-2), we have: f(2) = -2 Substitute x=2 into the function: a(2)^2 + b(2) + c = -2 This simplifies to: 4a + 2b + c = -2 This is our third equation.

Step 5 - Solve the system of equations

We have three equations with three unknowns: 1. a + b + c = 7 2. 2a + b = 0 3. 4a + 2b + c = -2 First, solve equation 2 for b: b = -2a Substitute b = -2a into equations 1 and 3: 1. a + (-2a) + c = 7 -a + c = 7 4a + 2(-2a) + c = -2 4a - 4a + c = -2 0 + c = -2 c = -2 Substitute c = -2 into -a + c = 7: -a - 2 = 7 -a = 9 a = -9 Finally, substitute a = -9 into b = -2a: b = -2(-9) b = 18

Key Concepts

Relative Maximum

A relative maximum is a point where a function reaches a peak within a specific interval. In this exercise, we need the quadratic function to have a relative maximum value of 7 when x=1.
To achieve this, we use two important aspects:
The value of the function at x=1 should be 7, which gives us the first equation: \(f(1) = 7\). By substituting, we get: \(a(1)^2 + b(1) + c = 7\), or \(a + b + c = 7\).
Additionally, the first derivative of the function must be zero at x=1. This means that the slope of the function is zero at the point, indicating a peak. The derivative of the function is \(f'(x) = 2ax + b\). Setting \(f'(1) = 0\) gives us the second equation: \(2a + b = 0\).
These two conditions help us confirm the relative maximum at the specified point.

System of Equations

A system of equations is a set of equations with multiple variables. To find the coefficients in our quadratic function, we need to solve a system of three equations.
We derived three equations from the conditions provided:

• From the relative maximum condition: \(a + b + c = 7\)
• From the derivative condition: \(2a + b = 0\)
• From the point condition that the function passes through (2,-2): \(4a + 2b + c = -2\)

With this system, we can solve for \(a\), \(b\), and \(c\) by substitution or elimination methods. First, solve one equation for a variable and substitute it into the others. For example, solving \(2a + b = 0\) for \(b\), we get \(b = -2a\). Replacing \(b\) in the other equations simplifies them and allows us to solve for \(a\) and \(c\), followed by \(b\).

First Derivative

The first derivative of a function describes its rate of change. For quadratic functions like \(f(x) = ax^2 + bx + c\), the first derivative is \(f'(x) = 2ax + b\).
To find a relative maximum, the first derivative is set to zero, resulting in a flat slope. We found that \(f'(1) = 0\), which means:\(2a(1) + b = 0\), or \(2a + b = 0\). This helps us identify one of our crucial equations to solve for the coefficients.
The first derivative is also useful in verifying whether the critical points are maxima or minima by using the second derivative test. However, in this specific exercise, due to the nature of the problem, we only used the first derivative to find crucial conditions for the quadratic function.