Problem 49 - Precalculus with Calculus Previews

Question

In Problems \(49-52\), find at least one function defined implicitly by the given equation. Give the domain of each function. \((x-1)^{2}=\frac{1}{2}(y+2)\)

Step-by-Step Solution

Answer

The function is \(y = 2(x-1)^2 - 2\) with domain \((- infty, infty)\).

Step 1: Rearrange the Equation

Start with the given equation \[(x-1)^2 = \frac{1}{2}(y+2)\]First, solve for \(y\) by multiplying both sides by 2. This gives:\[2(x-1)^2 = y+2\]

Step 2: Isolate y

Now, isolate \(y\) by subtracting 2 from both sides:\[y = 2(x-1)^2 - 2\]

Step 3: Identify the Function and Domain

The implicit function derived is \(y = 2(x-1)^2 - 2\). This is a quadratic function and is defined for all real numbers. Thus, the domain of the function is \( (-\infty, \infty) \).

Key Concepts

Quadratic Equations

In the realm of mathematics, quadratic equations are fundamental and widely encountered. A quadratic equation is typically one of the form \( ax^2 + bx + c = 0 \). This equation involves terms where the highest power of the variable, usually denoted by \( x \), is a square (hence the term "quadratic").

In the context of the exercise, we are dealing with a quadratic function, presented in a rearranged form as \( y = 2(x-1)^2 - 2 \). Here's what you should know about quadratic equations:

• Standard Form: A quadratic equation in its standard form is \( ax^2 + bx + c = y \). Depending on the values of \( a, b, \) and \( c \), the graph of the equation will be a parabola.
• Vertex Form: Our derived function \( y = 2(x-1)^2 - 2 \) is in vertex form, \( y = a(x-h)^2 + k \), which makes it easy to see that the vertex is at \((h, k) = (1, -2)\).
• Symmetry: Parabolas have an axis of symmetry at \( x = h \), which means they mirror themselves across this vertical line.

Understanding these basic characteristics of quadratic equations can significantly help in analyzing and graphing them. The equation we have here also opens upwards because the coefficient of \( (x-1)^2 \) is positive.

Domain of a Function

The "domain" of a function refers to all the possible input values \( x \) that the function can accept without encountering undefined expressions or other mathematical issues.

For the equation \( y = 2(x-1)^2 - 2 \), it is a quadratic function, which typically has a domain of all real numbers since there are no restrictions on the values \( x \) can take. To determine this, consider the following:

• Polynomials: Quadratic functions are polynomials. Polynomials, in general, are defined for all real number inputs.
• No Division or Roots: In the absence of division by the variable or root operations (as they could introduce restrictions like non-zero division or non-negative radicands), the function remains continuous and uninterrupted across the entire number line.

As a result, the domain of the function \( y = 2(x-1)^2 - 2 \) is \( (-\infty, \infty) \), meaning it is defined for any real number you can think of. This makes such functions especially easy to handle in many situations.

Function Isolation

Function isolation is the process of solving an equation for one variable in terms of another. This allows us to express one variable (like \( y \)) explicitly rather than implicitly.

In our exercise, the equation \( (x-1)^2 = \frac{1}{2}(y+2) \) was given implicitly. To isolate \( y \), we first manipulated the equation by performing operations on both sides:

• Multiply Both Sides: Start by multiplying through by 2 to clear the fraction: \( 2(x-1)^2 = y + 2 \).
• Subtract: Then, subtract 2 from both sides to solve for \( y \): \( y = 2(x-1)^2 - 2 \).

Isolating \( y \) helps us see it clearly as a function of \( x \) and provides us insight into the nature of the relationship between \( x \) and \( y \). It's like untangling a thread, making understanding and further analysis of the function much simpler.