Problem 12 - College Algebra

Question

For the following exercises, determine whether the relation represents \(y\) as a function of \(x\). $$ 2 x+y^{2}=6 $$

Step-by-Step Solution

Answer

No, it's not a function since \(y\) can take two values for each \(x\).

Step 1: Understand the Exercise

We need to determine if the equation \(2x + y^2 = 6\) represents \(y\) as a function of \(x\). This means that for each value of \(x\), there should be no more than one corresponding value of \(y\).

Step 2: Isolate y

Rewrite the equation to solve for \(y\): \[y^2 = 6 - 2x\]This expresses \(y^2\) in terms of \(x\).

Step 3: Examine the Expression for y

For \(y\) to be a function of \(x\), \(y\) must have a unique value for each \(x\). This requires solving for \(y\): \[y = \pm \sqrt{6 - 2x}\]Here, \(y\) can take two different values (\(+\sqrt{6-2x}\) and \(-\sqrt{6-2x}\)) for the same \(x\), indicating multiple possibilities.

Step 4: Conclusion

Since \(y\) is not uniquely determined by \(x\) (because it can take two values for each \(x\)), the given relation does not represent \(y\) as a function of \(x\).

Key Concepts

Relation Analysis

When we talk about relation analysis, we're diving into how two variables, like \( x \) and \( y \), interact via a mathematical relationship. In the equation \( 2x + y^2 = 6 \), this relationship isn't immediately clear until we apply some algebra.

Analysing this relation starts by trying to see if \( y \) can be expressed solely in terms of \( x \), and more specifically, if each value of \( x \) produces a single, unique \( y \).

• This is crucial for determining if the given equation is indeed a function.
• A function is essentially a special type of relation where there's only one output value for each input value.

Understanding whether a relation is a function helps us manage expectations about the possibilities that arise from inputting a value.

Solving Equations

To understand whether \( y \) is dependent solely on \( x \), we first need to solve the equation for \( y \). From our original equation, \( 2x + y^2 = 6 \), the goal is to isolate \( y \).

Subtract \( 2x \) from both sides, leading to the equation \( y^2 = 6 - 2x \). Now, attempt taking the square root to solve for \( y \):

• This gives us \( y = \pm \sqrt{6 - 2x} \).
• Notice the plus-minus (\( \pm \)), indicating two potential outputs for \( y \) whenever \( x \) is plugged in.

This dual solution shows the step-by-step unraveling of the equation's nature, highlighting why this specific equation doesn't classify \( y \) as a unique function of \( x \).

Unique Values

Understanding unique values in the context of functions is key. For the equation \( y = \pm \sqrt{6 - 2x} \), each \( x \) that we substitute into the equation could potentially give us two separate values for \( y \) because of the \( \pm \) symbol.

• When an equation is supposed to represent a function, every \( x \) should map to just one \( y \).
• If any \( x \) corresponds to multiple \( y \) values, it's a sure sign the relation is not a function.

This lack of uniqueness in values directly leads us to the conclusion, reinforcing that the given relation fails the function test.

Graphing Relations

Visualizing this equation with a graph can make the non-function nature of the relation evident. When we graph \( y = \pm \sqrt{6 - 2x} \), the picture becomes clear:

• The graph resembles a sideways parabola, stretching horizontally on the \( x \)-axis.
• This sideways curve crosses any given vertical line (vertical line test) at two points within its domain.

If a vertical line crosses the graphed curve more than once within the graph's domain, it proves the relation doesn't meet the criteria of a function.

Thus, graphing not only solidifies the algebraic analysis but also provides a visual proof that \( y \) is not a unique function of \( x \).