Problem 7 - Algebra 1

Question

Evaluate the function for \(x=0,1,2,3,\) and \(4 .\) Round your answer to the nearest tenth. $$ y=\sqrt{x+2} $$

Step-by-Step Solution

Answer

The respective values of the function for \(x = 0,1,2,3,4\) are 1.4, 1.7, 2.0, 2.2, 2.4 when rounded to the nearest tenth

Step 1: Evaluation for x=0

Substitute \(x = 0\) into the function. We get \(y = \sqrt{0+2} = \sqrt{2}\). This rounds to 1.4 when rounded to the nearest tenth

Step 2: Evaluation for x=1

Now substitute \(x = 1\). Therefore we get \(y = \sqrt{1+2} = \sqrt{3}\). This rounds to 1.7 when rounded to the nearest tenth

Step 3: Evaluation for x=2

Now, with \(x = 2\), the function becomes \(y = \sqrt{2+2} = \sqrt{4}\). This equals 2.0 when rounded to the nearest tenth

Step 4: Evaluation for x=3

With \(x = 3\), we get \(y = \sqrt{3+2} = \sqrt{5}\). This rounds to 2.2 when rounded to the nearest tenth

Step 5: Evaluation for x=4

Finally, when \(x = 4\), the function becomes \(y = \sqrt{4+2} = \sqrt{6}\). This rounds to 2.4 when rounded to the nearest tenth

Key Concepts

Square Root Function

A square root function involves finding a number that, when multiplied by itself, gives the original number under the square root sign. In other words, the square root of a number \( x \), denoted as \( \sqrt{x} \), is a value \( y \) such that \( y^2 = x \). These functions are usually expressed in equation form, where the output \( y \) is a result of solving \( \sqrt{x+c} \), making it a basic type of radical function that exhibits a half-parabolic shape when graphed.

If we take the given function \( y = \sqrt{x + 2} \), we can classify it as a shifted square root function. The expression under the square root, \( x + 2 \), implies that for every value of \( x \), you are adding 2 before finding the square root. This results in a slight horizontal shift in the graph of \( \sqrt{x} \).

The fascinating part of square root functions is that they are only defined for non-negative values of the expression inside the root, in this case, \( x + 2 \geq 0 \). This means that \( x \) must be greater than or equal to \(-2\) to yield real number outputs. In practice, however, we often stick to non-negative \( x \) values for ease of calculations.

Rounding Decimals

Rounding decimals is a crucial skill in math that simplifies numbers, making them easier to work with. It involves adjusting a decimal number to retain a specific degree of accuracy. The concept is especially useful in real-world calculations where exact precision might be unnecessary or cumbersome.

When rounding to the nearest tenth, you focus on the first decimal place. Here's how you can do it:

• Look at the number in the second decimal place.
• If this number is 5 or greater, you increase the first decimal place by one.
• If it is less than 5, you leave the first decimal place as is.

For example, rounding \( \sqrt{2} \approx 1.414 \) to the nearest tenth gives you 1.4, because the second decimal place, 1, is less than 5.

This method helps to communicate numbers more effectively by maintaining the level of accuracy needed for practical purposes, such as when measuring or calculating distances and quantities. Remember that while rounding makes numbers less precise, it also makes them much more digestible when presented.

Evaluating Expressions

Evaluating an expression involves substituting a given value for the variable and then performing the calculations needed to simplify the expression to a single numerical output. This is a fundamental skill in algebra and calculus, allowing you to understand the behavior of functions for specific inputs.

Using the original exercise as a reference, you evaluate the function \( y = \sqrt{x+2} \) by replacing \( x \) with a given number:

• For \( x = 0 \), substitute to get \( y = \sqrt{0+2} = \sqrt{2} \), which simplifies and rounds to 1.4.
• For \( x = 1 \), substitute to obtain \( y = \sqrt{1+2} = \sqrt{3} \), which rounds to 1.7.
• Continue this process for each value of \( x \), progressively solving the expression to find the corresponding \( y \) values.

Evaluating expressions requires attention to detail in substituting and simplifying, ensuring accurate results. As practice, it sharpens your mathematical thinking and helps grasp how different inputs affect the output of a function. This skill is foundational for more advanced topics and problem-solving in mathematics.