Problem 27 - Applied Calculus

Question

Find constants \(a\) and \(b\) so that the minimum for the parabola \(f(x)=x^{2}+a x+b\) is at the given point. [Hint: Begin by finding the critical point in terms of \(a\). $$ (3,5) $$

Step-by-Step Solution

Answer

The constants are \(a = -6\) and \(b = 14\).

Step 1: Identify the Objective for the Parabola

The goal is to find constants \(a\) and \(b\) such that the parabola \(f(x) = x^2 + ax + b\) has its vertex (minimum point) at the given point \((3, 5)\).

Step 2: Determine the Vertex Formula

For a parabola \(f(x) = x^2 + ax + b\), the x-coordinate of the vertex is given by \(-\frac{a}{2}\). We are given that the x-coordinate of the vertex is 3. Set \(-\frac{a}{2} = 3\).

Step 3: Solve for \(a\)

From \(-\frac{a}{2} = 3\), solve for \(a\):\[-a = 6 \rightarrow a = -6.\]

Step 4: Use the Vertex to Find \(b\)

Now that \(a = -6\), substitute \(x = 3\) into the equation \(f(x) = x^2 + ax + b\) and use \(f(3) = 5\) to find \(b\):\[f(3) = 3^2 - 6 \cdot 3 + b = 5.\]Simplify:\[9 - 18 + b = 5 \rightarrow -9 + b = 5.\]

Step 5: Solve for \(b\)

From \(-9 + b = 5\), solve for \(b\):\[ b = 5 + 9 \rightarrow b = 14. \]

Key Concepts

Parabola Vertex

A parabola is a U-shaped graph represented by a quadratic function of the form \( f(x) = ax^2 + bx + c \) or similar variations. One of its key features is the vertex, which is the highest or lowest point on the parabola. The vertex helps us understand many properties of the graph, such as its direction and maximum or minimum values.
In our given equation \( f(x) = x^2 + ax + b \), the vertex can be found using the formula for the x-coordinate:

• The x-coordinate of the vertex is given by the formula: \( x = -\frac{a}{2} \).
• By plugging the appropriate value of x into the function, we can find the corresponding y-coordinate.

In this exercise, we are given that the vertex is \((3, 5)\), meaning the x-coordinate of the vertex is 3. Thus, using \(-\frac{a}{2} = 3\), we find that \(a\) is -6. This calculation allows us to define the position of the parabola on the graph accurately.

Minimum Point

The minimum point of a parabola occurs at its vertex when the parabola is oriented upwards (when the quadratic term's coefficient is positive). For \(f(x) = x^2 + ax + b\), the quadratic term is \(x^2\), having a coefficient of 1, meaning our parabola opens upwards.
Finding the minimum point means determining both x and y coordinates of the vertex:

• Use \(-\frac{a}{2} = x\) to find the x-coordinate; once found, substitute into \(f(x)\) to find y.
• In this exercise, the given minimum point is \((3, 5)\), which means at \(x = 3\), \(f(3) = 5\).

Determining these coordinates helps in understanding the nature of problems involving optimization and provides crucial insight into the range and behavior of quadratic functions. This is valuable when modeling real-world scenarios that require minimum values, such as cost minimization.

Critical Point Calculation

Critical points are places on a graph where a function's slope is zero, often corresponding to local maximum or minimum points. For a quadratic function, these critical points coincide with the vertex of the parabola.

In our equation \( f(x) = x^2 + ax + b \), the critical point is where the derivative equals zero: \(f'(x) = 0\).

• The derivative of \( f(x) = x^2 + ax + b \) is \( f'(x) = 2x + a \).
• Set \( f'(x) = 0 \) to find: \(2x + a = 0\). Solving for \(x\) gives the x-coordinate of the critical point.

For this function, the critical calculation confirms what was found in previous steps for finding \(a\) by setting \(-\frac{a}{2} = 3\). This reaffirms the critical point as the vertex \((3, 5)\). Critical point calculations are essential for correctly identifying pivotal points where functions change their increasing or decreasing behavior, a key concept in calculus and optimization.