Problem 50 - Precalculus Mathematics for Calculus

Question

Determine whether the given points are on the graph of the equation. $$y\left(x^{2}+1\right)=1 ; \quad(1,1),\left(1, \frac{1}{2}\right),\left(-1, \frac{1}{2}\right)$$

Step-by-Step Solution

Answer

Points (1, 1/2) and (-1, 1/2) are on the graph.

Step 1: Understand the Equation

The given equation is \( y(x^2 + 1) = 1 \). We need to check if each point satisfies this equation.

Step 2: Test the Point (1, 1)

Substitute \( x = 1 \) and \( y = 1 \) into the equation. The equation becomes \( 1(1^2 + 1) = 1 \). Simplifying, we have \( 1(1 + 1) = 2 \), which is not equal to 1. Thus, (1, 1) is not on the graph.

Step 3: Test the Point (1, 1/2)

Substitute \( x = 1 \) and \( y = \frac{1}{2} \) into the equation. The equation becomes \( \frac{1}{2}(1^2 + 1) = 1 \). Simplifying, \( \frac{1}{2}(2) = 1 \), which is equal to 1. Thus, (1, 1/2) is on the graph.

Step 4: Test the Point (-1, 1/2)

Substitute \( x = -1 \) and \( y = \frac{1}{2} \) into the equation. The equation becomes \( \frac{1}{2}((-1)^2 + 1) = 1 \). Simplifying, \( \frac{1}{2}(1 + 1) = 1 \) or \( \frac{1}{2}(2) = 1 \), which is also equal to 1. Thus, (-1, 1/2) is on the graph.

Key Concepts

Points on a Graph

Imagine the graph as a map and each point as a location or address on this map. When we say a point is "on the graph," it means this location precisely aligns with where the instructions (equation) say it should be.
For this task, the equation given is \( y(x^2 + 1) = 1 \). To determine if a point like \((1, 1)\) or \((1, \frac{1}{2})\) lies on the graph, we substitute these values into the equation and check if the equation holds true.
If the left-hand side of the equation equals the right-hand side after substitution, that point is exactly "on" the graph. Otherwise, it's off the map, so to speak.

• Each point is a pair of numbers – \((x, y)\).
• Take the \(x\) coordinate to work within the equation.
• Match the result with the \(y\) value to see if the equation is valid.

Understanding this helps us visualize what solving an equation means: confirming if certain positions meet the key condition of the given equation.

Substitution Method

The substitution method is like peeling an onion, where each layer gives us more insight. To verify if a point lies on the graph, substitution is our tool. Here's how it works: take each pair \((x, y)\), and put these values into the equation to see if it fits.
In the example with \( y(x^2 + 1) = 1 \), substituting \((1, 1)\) asks us to replace \(x\) with 1 and \(y\) with 1. What makes this helpful is that it turns an abstract equation into simple arithmetic.

• Begin with taking the \(x\) value from the point and replace it in the equation.
• Follow this by inserting the \(y\) value as well, completing the equation’s blanks.
• Simplify the equation.

The key is in seeing the equation as a test. If everything computes correctly on both sides, the point is on the graph. Simple substitutions transform a complex chase for values into straightforward numbers work.
This structured approach turns your question into straightforward steps.

Checking Solutions for Equations

Checking solutions is the detective work behind math problems. It ensures that each solution is spot on. To affirm if a point is a solution to the equation, we review if it satisfies the equation meaningfully.
This involves making sure both sides of the equation balance once the point is substituted. The point \((1, \frac{1}{2})\), for instance, checks out because substituting \(x = 1\) and \(y = \frac{1}{2}\) results in the left side also being 1, making the equation true: \(\frac{1}{2}(2) = 1\) is correct.
Steps include:

• Substitute \(x\) and \(y\) from your point into the equation step-by-step.
• Ensure all calculations are accurate.
• If both sides of the equation equal, the solution is correct.

This process is like cross-checking your work in an assignment. If the math balances, the solution is indeed valid, proving that knowledge with certainty and confidence.