Problem 28 - 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 \(a\). $$ (-2,-3) $$

Step-by-Step Solution

Answer

The constants are \(a = 4\) and \(b = 1\).

Step 1: Identify the Function

The given function is a parabola, represented by \( f(x) = x^2 + ax + b \). We need to adjust the constants \(a\) and \(b\) so that its minimum occurs at the point \((-2, -3)\).

Step 2: Find the Vertex Formula

The formula for the vertex of a parabola given by \( f(x) = ax^2 + bx + c \) is \( x = -\frac{b}{2a} \). For our function, \( a = 1 \), so the x-coordinate of the vertex is \( x = -\frac{a}{2}\).

Step 3: Set the Vertex x-coordinate

We need the x-coordinate of the vertex to be \(-2\). Thus, we have:\[ -\frac{a}{2} = -2\]

Step 4: Solve for Constant \(a\)

From the equation in Step 3, solve for \(a\):\[ -\frac{a}{2} = -2 \Rightarrow a = 4\]

Step 5: Use the Given Point to Solve for \(b\)

We know that the minimum value of the function \(f\) at \(x = -2\) should be \(-3\). Substitute \(x = -2\) and \(f(x) = -3\) into the function:\[ (-2)^2 + 4(-2) + b = -3\]Simplify to solve for \(b\).

Step 6: Solve for Constant \(b\)

The equation from Step 5 becomes:\[ 4 - 8 + b = -3\]\[ b - 4 = -3\]\[ b = 1\]

Step 7: Verify the Result

Substituting \(a = 4\) and \(b = 1\) back into the function:\[ f(x) = x^2 + 4x + 1 \]Calculate \( f(-2) \) to verify it yields \(-3\):\[ (-2)^2 + 4(-2) + 1 = 4 - 8 + 1 = -3\]This confirms the solution.

Key Concepts

Vertex Calculation

Calculating the vertex of a parabola is an important step in understanding and manipulating its graph. The vertex is the point where the parabola changes direction, typically representing a minimum or maximum value, depending on the direction of the parabola.For a quadratic function in the standard form, given by the equation:\[ f(x) = ax^2 + bx + c \]The vertex can be found using the vertex formula for the x-coordinate:\[ x = -\frac{b}{2a} \]In this specific problem, the quadratic function is adjusted to ensure the minimum occurs at a specific point. Initially, the function is given as \( f(x) = x^2 + ax + b \), where the coefficient of \( x^2 \) is \( 1 \), so:\[ x = -\frac{a}{2} \]To make the vertex occur at \( x = -2 \), you solve:\[ -\frac{a}{2} = -2 \]This calculation leads to \( a = 4 \). Therefore, modifying the function to move the vertex to the desired location requires calculating the appropriate \( a \) value.

Critical Points

Critical points in a quadratic function are values of \( x \) at which the function reaches a local extremum (either a minimum or maximum). For parabolas, these points are easily identifiable because they correspond to the vertex of the function.A critical point for the quadratic function:\[ f(x) = x^2 + ax + b \]occurs when the derivative of the function is zero. The derivative, in terms of \( x \), is:\[ f'(x) = 2x + a \]Setting \( f'(x) = 0 \) gives us the critical point:\[ 2x + a = 0 \]For this exercise, we're interested in making sure the minimum occurs at \( x = -2 \). With the derivative set to zero, this leads to:\[ 2(-2) + a = 0 \]Solving this ensures that the critical point aligns with the calculated vertex, thus confirming that \( a = 4 \). Therefore, the chosen vertex is indeed a critical point.

Quadratic Functions

Quadratic functions are fundamental in algebra and form the basis of parabolas. They are expressed in the standard format:\[ f(x) = ax^2 + bx + c \]where \( a \), \( b \), and \( c \) are constants with \( a eq 0 \). These functions graph into a U-shaped curve known as a parabola. Depending on the sign of \( a \), the parabola opens upwards (\( a > 0 \)) or downwards (\( a < 0 \)).In this problem, the given quadratic is:\[ f(x) = x^2 + ax + b \]Here, \( a \) affects the slope and direction of the \( x \) term, while \( b \) shifts the parabola vertically. Calculating specific constants \( a \) and \( b \) to ensure the parabola has a vertex at \((-2, -3)\) involves:

• Determining \( a \) by aligning the vertex formula \( x = -\frac{a}{2} \) with the given \( x \)-coordinate.
• Solving for \( b \) by using the vertex point \((-2, -3)\) to ensure the function hits the minimum value at \( x = -2 \).

Ultimately, quadratic functions, while simple, encourage exploration of algebraic operations and are critical for learning how different algebraic forms influence graphical behavior.