Question

Evaluate the function for \(x=0,1,2,3,\) and \(4 .\) Round your answer to the nearest tenth. $$ y=6 \sqrt{x}-3 $$

Short answer

The values of \(y\) for \(x = 0, 1, 2, 3, 4\) are -3.0, 3.0, 5.5, 7.4, and 9.0 respectively.

Step-by-step solution

Substitution for \(x = 0\)

Substitute \(x = 0\) into the function: \(y = 6 * \sqrt{0} - 3 = -3\). After rounding it to the nearest tenth, it remains -3.0

Substitution for \(x = 1\)

Substitute \(x = 1\) into the function: \(y = 6 * \sqrt{1} - 3 = 6 - 3 = 3\). After rounding it to the nearest tenth, it remains 3.0

Substitution for \(x = 2\)

Substitute \(x = 2\) into the function: \(y = 6 * \sqrt{2} - 3\). Numerically, this equals approximately 5.49. After rounding it to the nearest tenth, the result is 5.5

Substitution for \(x = 3\)

Substitute \(x = 3\) into the function: \(y = 6 * \sqrt{3} - 3\). Numerically, this equals approximately 7.40. After rounding it to the nearest tenth, the result is 7.4

Substitution for \(x = 4\)

Substitute \(x = 4\) into the function: \(y = 6 * \sqrt{4} - 3 = 9\). After rounding it to the nearest tenth, it remains 9.0

Key concepts

Square Root Operations

Understanding square root operations is essential when evaluating functions that involve square roots. The square root of a number is a value that, when multiplied by itself, gives the original number. For example, the square root of 9 is 3, since 3 times 3 is 9.

When dealing with square roots in algebra, it's important to remember that they represent the principal, or non-negative root. For instance, even though both 3 and -3 squared equal 9, the square root of 9 is defined as the positive value 3. This becomes particularly important in evaluating functions where precise positive values are expected as a result.

In our exercise involving the function \(y=6 \sqrt{x}-3\), calculations are simplified by first finding the square root of the given \(x\) value. The square roots of whole numbers like 0, 1, and 4 are simple, as they return 0, 1, and 2, respectively. However, square roots of non-perfect squares, such as 2 and 3, are irrational numbers. They cannot be expressed as a precise fraction or decimal, which is why we need to approximate them before we can proceed with further operations or rounding.

Substitution Method

The substitution method is a key algebraic technique used to evaluate functions. It involves replacing a variable in an equation with its corresponding value. It's a straightforward practice: wherever you see the variable, you replace it with the given number or expression.

In the context of the given exercise, we are asked to evaluate the function \(y=6 \sqrt{x}-3\) for various values of \(x\) - specifically 0, 1, 2, 3, and 4. To do this, we systematically substitute each of these values in place of the variable \(x\) and then carry out the required operations.

For instance, when \(x=2\), we replace \(x\) in the function with 2 to get \(y=6 \sqrt{2}-3\). The substitution method is crucial in understanding how functions behave for different inputs, which helps in grasping the relationship between the variable \(x\) and the outcome of the function \(y\).

Rounding Numbers

Rounding numbers is an essential numerical skill, especially when dealing with non-whole numbers or irrational numbers. The process involves approximating a number to the nearest specified place value – often to the nearest whole number, tenth, hundredth, and so on, depending on the level of precision required.

In the exercise, after evaluating the function formally, we are told to round our answers to the nearest tenth. This means we are considering numbers to one decimal place. For example, after substitution and calculation for \(x=2\), we got approximately 5.49. To round to the nearest tenth, we look at the number in the hundredths place (9 in this case) and round up the tenth's place, resulting in 5.5.

The ability to round numbers is especially vital when dealing with irrational numbers or lengthy decimals, as it allows us to work with more manageable numbers without significantly compromising accuracy. It's a technique that's not only useful in pure mathematics but also widely applicable in real-world contexts where estimations are required.