Problem 83 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the value of \(b\) such that the function has the given maximum or minimum value. \(f(x)=x^{2}+b x+26 ;\) Minimum value: 10

Step-by-Step Solution

Answer

The values of \(b\) that make the function \(f(x)=x^2+bx+26\) have a minimum value of 10 are \(b = 8\) and \(b=-8\).

Step 1: Find the x-coordinate of vertex

Given the function \(f(x)=x^2+bx+26\), we find the x-coordinate \(h\) of the vertex using the formula \(-\frac{b}{2a}\). Here, \(a=1\) and \(b=b\), so \(h=-\frac{b}{2(1)}= -\frac{b}{2}\)

Step 2: Apply the x-coordinate in the function

Now, plugging \(h=-\frac{b}{2}\) into the function \(f(x)\) to find the y-coordinate \(k\), we get: \(k=f(h)=\left(-\frac{b}{2}\right)^2 + b\left(-\frac{b}{2}\right) + 26= \frac{b^2}{4} - \frac{b^2}{2} +26 = \frac{b^2}{4} - \frac{2b^2}{4} + 26 = \frac{-b^2}{4} +26\)

Step 3: Set the y-coordinate equal to the minimum value and solve for \(b\)

The minimum value of function \(f(x)\) is given as 10. We set the y-coordinate equal to 10 and solve for \(b\): \(10 = \frac{-b^2}{4} +26\). This simplifies to \(b^2 = 4(26-10)=4*16=64\). Taking the square root of both sides, remembering that it could be either positive or negative, we get \(b=\pm 8\)

Key Concepts

Vertex Form of a Quadratic Function

The vertex form of a quadratic function is a convenient way to represent a quadratic equation. It is particularly useful when you are interested in identifying the vertex of the parabola represented by the quadratic function. The standard form of a quadratic function is given by:

• \( f(x) = ax^2 + bx + c \)

To convert it to the vertex form, we use the expression:

• \( f(x) = a(x-h)^2 + k \)

Here,

• \( (h, k) \) represents the vertex of the parabola.
• \( h \) is the x-coordinate, and \( k \) is the y-coordinate of the vertex.
• \( a \) determines the direction and width of the parabola.

To find the vertex coordinates from the standard form, we use the formula \( h = -\frac{b}{2a} \). Once \( h \) is known, plug it back into the original quadratic equation to find \( k \). This simple transformation provides significant insights into the shape and position of the parabola, making it easier to analyze.

• Understanding the vertex helps identify the maximum or minimum value of the function.

Minimum Value of a Function

In quadratic functions like \( f(x) = ax^2 + bx + c \), the minimum value is directly related to the vertex. When \( a > 0 \), the parabola opens upwards, and the vertex represents the minimum point.
Finding the minimum involves two main steps. First, find the x-coordinate of the vertex using \( h = -\frac{b}{2a} \). Then, substitute the \( h \) value back into the function to determine the y-coordinate, which is the minimum value of the function.
This minimum value, given by \( k \) in vertex form, is significant in various contexts:

• It helps understand real-world problems where minimizing cost or maximizing efficiency is essential.
• It provides crucial points in graphing the quadratic function.

Solving Quadratic Equations

Solving quadratic equations involves finding values of the variable that satisfy the equation. These are known as the roots or solutions of the quadratic equation. For a standard form \( ax^2 + bx + c = 0 \), there are several methods to solve it:

• Factoring: Express the quadratic as a product of two binomials, if possible.
• Quadratic Formula: Apply \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) when factoring is inconvenient.
• Completing the Square: Rewriting the equation to achieve a perfect square trinomial.

Each method has its own advantages depending on the context and complexity of the equation. In the given original step-by-step solution, after deriving the expression for the vertex, we find the parameters such as \( b \) by setting the minimum value and solving the resulting equation. By equating the expression to the minimum value, solving for \( b \) yields multiple solutions: \( b = \pm 8 \).

• This shows the symmetrical property of parabolas, which often results in two potential values for certain parameters.