Analyzing the Function Options

Oct 13, 2025 by talynt.org 53 views

In mathematics, functions are used to describe the relationship between input and output values. A function is a rule that assigns to each input value, called the domain, a unique output value, called the range. In this article, we will explore a function named g $}$ with an input of <span class="katex-error" title="ParseError KaTeX parse error: Expected '', got 'EOF' at end of input: " style="color#cc0000">{ c $$ and a range of all real numbers.

Analyzing the Function Options

We are given two options to represent the function ${$ g $}$:

A. ${$ c(g) = \frac{4}{5} g + \frac{2}{5} $}$

B. ${$ g(c) = 4c^2 - 2c + 1 $}$

To determine which option best represents the described function, we need to analyze each option carefully.

Option A: ${$ c(g) = \frac{4}{5} g + \frac{2}{5} $}$

In this option, the function is represented as ${$ c(g) = \frac{4}{5} g + \frac{2}{5} $}$. This means that the input value ${$ c $}$ is being multiplied by ${$ \frac{4}{5} $}$ and then added to ${$ \frac{2}{5} $}$. However, this option does not match the description of the function, which has an input of ${$ c $}$ and a range of all real numbers.

Option B: ${$ g(c) = 4c^2 - 2c + 1 $}$

In this option, the function is represented as ${$ g(c) = 4c^2 - 2c + 1 $}$. This means that the input value ${$ c $}$ is being squared and multiplied by 4, then subtracted by 2 times ${$ c $}$, and finally added to 1. This option matches the description of the function, which has an input of ${$ c $}$ and a range of all real numbers.

Conclusion

Based on the analysis of the two options, we can conclude that the best representation of the function ${$ g $}$ is option B: ${$ g(c) = 4c^2 - 2c + 1 $}$. This option matches the description of the function, which has an input of ${$ c $}$ and a range of all real numbers.

Understanding the Function Graph

To visualize the function, we can graph the function using a coordinate plane. The x-axis represents the input value ${$ c $}$, and the y-axis represents the output value ${$ g(c) $}$. The graph of the function will be a parabola that opens upwards, since the coefficient of the squared term is positive.

Graphing the Function

To graph the function, we can start by finding the vertex of the parabola. The vertex of the parabola is the point where the parabola changes direction. To find the vertex, we can use the formula:

${$ x = \frac{-b}{2a} $}$

In this case, ${$ a = 4 $}$ and ${$ b = -2 $}$. Plugging these values into the formula, we get:

${$ x = \frac{-(-2)}{2(4)} = \frac{2}{8} = \frac{1}{4} $}$

This means that the vertex of the parabola is at the point ${$ (\frac{1}{4}, g(\frac{1}{4})) $}$. To find the value of ${$ g(\frac{1}{4}) $}$, we can plug ${$ c = \frac{1}{4} $}$ into the function:

${$ g(\frac{1}{4}) = 4(\frac{1}{4})^2 - 2(\frac{1}{4}) + 1 = 4(\frac{1}{16}) - \frac{1}{2} + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{1}{4} - \frac{2}{4} + \frac{4}{4} = \frac{3}{4} $}$

This means that the vertex of the parabola is at the point ${$ (\frac{1}{4}, \frac{3}{4}) $}$.

Graphing the Parabola

To graph the parabola, we can start by plotting the vertex at the point ${$ (\frac{1}{4}, \frac{3}{4}) $}$. Then, we can plot two points on either side of the vertex. For example, we can plot the point ${$ (0, g(0)) $}$ and the point ${$ (1, g(1)) $}$. To find the value of ${$ g(0) $}$, we can plug ${$ c = 0 $}$ into the function:

${$ g(0) = 4(0)^2 - 2(0) + 1 = 1 $}$

This means that the point ${$ (0, 1) $}$ is on the graph of the function.

To find the value of ${$ g(1) $}$, we can plug ${$ c = 1 $}$ into the function:

${$ g(1) = 4(1)^2 - 2(1) + 1 = 4 - 2 + 1 = 3 $}$

This means that the point ${$ (1, 3) $}$ is on the graph of the function.

Conclusion

In the previous article, we explored a function named ${$ g $}$ with an input of ${$ c $}$ and a range of all real numbers. We analyzed two options to represent the function and concluded that the best representation is option B: ${$ g(c) = 4c^2 - 2c + 1 $}$. In this article, we will answer some frequently asked questions about the function representation.

Q: What is the domain of the function?

A: The domain of the function is all real numbers, which means that the input value ${$ c $}$ can be any real number.

Q: What is the range of the function?

A: The range of the function is all real numbers, which means that the output value ${$ g(c) $}$ can be any real number.

Q: Is the function linear or non-linear?

A: The function is non-linear, since it involves a squared term ${$ c^2 $}$.

Q: Can the function be represented in a different form?

A: Yes, the function can be represented in a different form. For example, it can be represented as ${$ g(c) = (2c - 1)^2 $}$.

Q: How can I graph the function?

A: To graph the function, you can start by plotting the vertex at the point ${$ (\frac{1}{4}, \frac{3}{4}) $}$. Then, you can plot two points on either side of the vertex. For example, you can plot the point ${$ (0, g(0)) $}$ and the point ${$ (1, g(1)) $}$.

Q: What is the vertex of the parabola?

A: The vertex of the parabola is the point ${$ (\frac{1}{4}, \frac{3}{4}) $}$.

Q: How can I find the value of ${$ g(c) $}$ for a given value of ${$ c $}$?

A: To find the value of ${$ g(c) $}$ for a given value of ${$ c $}$, you can plug the value of ${$ c $}$ into the function ${$ g(c) = 4c^2 - 2c + 1 $}$.

Q: Can I use the function to model real-world phenomena?

A: Yes, the function can be used to model real-world phenomena. For example, it can be used to model the growth of a population over time.

Q: How can I use the function to solve problems?

A: To use the function to solve problems, you can plug in values of ${$ c $}$ and find the corresponding values of ${$ g(c) $}$. Then, you can use the values of ${$ g(c) $}$ to solve the problem.

Conclusion

In conclusion, the function ${$ g(c) = 4c^2 - 2c + 1 $}$ is a non-linear function with a domain and range of all real numbers. It can be graphed using a parabola with a vertex at the point ${$ (\frac{1}{4}, \frac{3}{4}) $}$. The function can be used to model real-world phenomena and solve problems.