Problem 6 - Algebra 1: Concepts and Skills

Question

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

Step-by-Step Solution

Answer

The evaluated function values are: for \(x=0\), \(y=-3.0\), for \(x=1\), \(y=3.0\), for \(x=2\), \(y=5.5\), for \(x=3\), \(y=6.4\), and for \(x=4\), \(y=9.0\).

Step 1 Evaluate for \(x=0\)

To begin, substitute \(x=0\) into the function, so \(y=6 \sqrt{0} - 3\). Evaluating the square root and performing the multiplication results in \(y=-3\). Rounding this to the nearest tenth, we get \(y=-3.0\)

Step 2 Evaluate for \(x=1\)

Next, substitute \(x=1\) into the function, so \(y=6 \sqrt{1} - 3\). Evaluating the square root and performing the multiplication results in \(y=3\). Rounding to the nearest tenth, we get \(y=3.0\)

Step 3 Evaluate for \(x=2\)

Now, substitute \(x=2\) into the function, so \(y=6 \sqrt{2} - 3\). After calculation, it gives approximately \(y=5.5\) when rounded to the nearest tenth.

Step 4 Evaluate for \(x=3\)

Next, substitute \(x=3\) into the function, so \(y=6 \sqrt{3} - 3\). After calculation, it gives approximately \(y=6.4\) when rounded to the nearest tenth.

Step 5 Evaluate for \(x=4\)

Finally, substitute \(x=4\) into the function, so \(y=6 \sqrt{4} - 3\). Evaluating the square root and performing the multiplication results in \(y=9\). Rounding to the nearest tenth, we get \(y=9.0\).

Key Concepts

Understanding Square Roots

The square root is a mathematical operation that is essential in many functions, including the one we're evaluating here. When you see the symbol \(\sqrt{x}\), it means you are looking for a number which, when multiplied by itself, gives you \(x\). For example, \(\sqrt{9} = 3\) because \(3 \times 3 = 9\). In our function, we had to find square roots of numbers like \(0\), \(1\), \(2\), \(3\), and \(4\).

• \(\sqrt{0} = 0\)
• \(\sqrt{1} = 1\)
• \(\sqrt{4} = 2\)

For other numbers like \(2\) and \(3\), the square roots are not whole numbers. For example, \(\sqrt{2} \approx 1.414\) and \(\sqrt{3} \approx 1.732\). The beauty of square roots comes in their ability to help solve real-world problems, such as finding the side length of a square with a given area.

Mastering Rounding Numbers

Rounding numbers is a handy technique when you want to simplify a number but still keep it close to the original value. It is especially useful when dealing with decimals in your calculations. The rule of rounding to the nearest tenth is straightforward: look at the hundredths place.

• If it's 5 or more, round the tenths place up.
• If it's less than 5, keep the tenths place the same.

For instance, when evaluating the function with \(x = 2\), the calculation yielded approximately \(5.485\). Since the number in the hundredths place (8) is more than 5, we round the result up to \(5.5\). Similarly, for \(x = 3\), the result was approximately \(6.372\), and since 7 is more than 5, we round it up to \(6.4\). Rounding simplifies calculations and often helps with reporting data in a clearer manner.

Function Evaluation Made Simple

Evaluating a function involves substituting the given input values into the function's formula and carrying out the necessary operations. Here, with the function \(y = 6\sqrt{x} - 3\), we plugged in values for \(x\) such as \(0\), \(1\), \(2\), \(3\), and \(4\).

• For \(x = 0\), the function becomes \(y = 6 \times 0 - 3\), resulting in \(y = -3\).
• For \(x = 1\), it simplifies to \(y = 6 \times 1 - 3\), giving \(y = 3\).
• For \(x = 4\), it’s \(y = 6 \times 2 - 3\), which yields \(y = 9\).

Function evaluation helps you understand how inputs and operations within a function translate to outputs. This process is useful in so many areas, from engineering to economics, allowing for better predictions and understanding of complex systems. By mastering evaluation, you’re essentially gaining a key to understanding how many real-world relationships work.