Problem 52 - Precalculus with Calculus Previews

Question

Find at least one function defined implicitly by the given equation. Give the domain of each function. \(y^{2}-12 y=x-11\)

Step-by-Step Solution

Answer

Functions: \( y = 6 \pm \frac{\sqrt{4x + 100}}{2} \); Domain: \( x \geq -25 \).

Step 1: Rearrange the given equation

To find a function defined implicitly by the given equation \( y^2 - 12y = x - 11 \), start by rearranging the equation to make it explicitly dependent on \( y \). This gives us the form: \( y^2 - 12y - x + 11 = 0 \).

Step 2: Recognize the structure of the equation

The equation is a quadratic in \( y \). In standard quadratic form, it is \( ay^2 + by + c = 0 \). Here, \( a = 1 \), \( b = -12 \), and \( c = -(x-11) \).

Step 3: Use the quadratic formula

We use the quadratic formula to solve for \( y \): \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \( a = 1 \), \( b = -12 \), and \( c = -(x-11) \).

Step 4: Substitute values and simplify

Substituting into the quadratic formula gives: \[ y = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot (-(x-11))}}{2 \cdot 1} = \frac{12 \pm \sqrt{144 + 4x - 44}}{2} \]. Simplify under the square root: \( \sqrt{4x + 100} \).

Step 5: Determine the functions

The solutions are: \[ y = \frac{12 + \sqrt{4x + 100}}{2} \] and \[ y = \frac{12 - \sqrt{4x + 100}}{2} \].

Step 6: Simplify the expressions

Simplify each expression: For \( y = \frac{12 + \sqrt{4x + 100}}{2} \), \( y = 6 + \frac{\sqrt{4x + 100}}{2} \). Similarly, for \( y = \frac{12 - \sqrt{4x + 100}}{2} \), \( y = 6 - \frac{\sqrt{4x + 100}}{2} \).

Step 7: Find the domain of each function

The term \( \sqrt{4x + 100} \) implies that \( 4x + 100 \geq 0 \), leading to \( x \geq -25 \). Thus, the domain for both functions is \( x \geq -25 \).

Key Concepts

Quadratic Equation

A quadratic equation is a fundamental mathematical expression that describes a parabolic curve. It is expressed in the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, and \( a eq 0 \). Here, "quadratic" refers to the highest degree of the variable being 2, which is represented by \( x^2 \).

When solving quadratic equations, the solutions can tell us where the parabola intersects the x-axis. These intersection points are also called the roots or zeros of the equation. Quadratic equations can have:

• Two real and distinct solutions
• One real solution (a repeated root)
• No real solutions (complex or imaginary roots)

For the given equation \( y^2 - 12y = x - 11 \), we recognize it is a quadratic in \( y \) by rearranging it to form \( y^2 - 12y - x + 11 = 0 \). Here \( a = 1, b = -12, \) and \( c = -(x - 11) \). Each value influences the nature of the roots, impacting how the equation can be solved and the behavior of the graph.

Quadratic Formula

The quadratic formula provides a straightforward method for finding the roots of any quadratic equation. It is written as \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula derives from completing the square on the standard quadratic equation \( ax^2 + bx + c = 0 \).

In our problem, this formula allows us to find \( y \) in terms of \( x \) efficiently by substituting \( a = 1 \), \( b = -12 \), and \( c = -(x - 11) \). What makes the quadratic formula particularly useful is its ability to work for any set of real numbers, revealing whether there are two solutions, one, or none, by evaluating the discriminant \( b^2 - 4ac \).

The discriminant determines the nature of the roots:

• If \( b^2 - 4ac > 0 \), two real and distinct solutions exist.
• If \( b^2 - 4ac = 0 \), there is exactly one real solution.
• If \( b^2 - 4ac < 0 \), the solutions are complex.

For the given equation, the expression under the square root simplifies to \( \sqrt{4x + 100} \), indicating that the solution involves two real roots as long as \( 4x + 100 \geq 0 \).

Domain of a Function

The domain of a function refers to all possible input values that will produce valid outputs. For most functions, it means identifying all values of \( x \) that do not result in undefined or non-real expressions.

In the context of functions involving square roots, such as \( \sqrt{4x + 100} \), the expression under the square root must be non-negative. Therefore, we set the inequality \( 4x + 100 \geq 0 \) to find the domain. Solving this gives \( x \geq -25 \), meaning \( x \) should be greater than or equal to \(-25\) to ensure the expression is defined and real.

Thus, for both functions derived from the equation \( y = 6 + \frac{\sqrt{4x + 100}}{2} \) and \( y = 6 - \frac{\sqrt{4x + 100}}{2} \), the domain is restricted to \( x \geq -25 \). Keeping track of the domain is essential as it outlines the valid input region for which the functions are applicable and behave as expected.