Evaluating the Function at Specific Values of x

Oct 13, 2025 by wbrum.org 25 views

Introduction

In this article, we will explore the concept of evaluating a quadratic function at specific values of x. A quadratic function is a polynomial function of degree two, which means the highest power of the variable (in this case, x) is two. The general form of a quadratic function is f(x) = ax^2 + bx + c, where a, b, and c are constants.

The Given Function

The given function is f(x) = 4x^2 + 8x - 5. This is a quadratic function with a = 4, b = 8, and c = -5.

Evaluating the Function at Specific Values of x

1. Evaluating f(0)

To evaluate f(0), we need to substitute x = 0 into the function f(x) = 4x^2 + 8x - 5.

f(0) = 4(0)^2 + 8(0) - 5 f(0) = 0 + 0 - 5 f(0) = -5

Therefore, f(0) = -5.

2. Evaluating f(2)

To evaluate f(2), we need to substitute x = 2 into the function f(x) = 4x^2 + 8x - 5.

f(2) = 4(2)^2 + 8(2) - 5 f(2) = 4(4) + 16 - 5 f(2) = 16 + 16 - 5 f(2) = 32 - 5 f(2) = 27

Therefore, f(2) = 27.

3. Evaluating f(-2)

To evaluate f(-2), we need to substitute x = -2 into the function f(x) = 4x^2 + 8x - 5.

f(-2) = 4(-2)^2 + 8(-2) - 5 f(-2) = 4(4) - 16 - 5 f(-2) = 16 - 16 - 5 f(-2) = 0 - 5 f(-2) = -5

Therefore, f(-2) = -5.

4. Evaluating f(x+1)

To evaluate f(x+1), we need to substitute x+1 into the function f(x) = 4x^2 + 8x - 5.

f(x+1) = 4(x+1)^2 + 8(x+1) - 5 f(x+1) = 4(x^2 + 2x + 1) + 8x + 8 - 5 f(x+1) = 4x^2 + 8x + 4 + 8x + 8 - 5 f(x+1) = 4x^2 + 16x + 7

Therefore, f(x+1) = 4x^2 + 16x + 7.

5. Evaluating f(x-1)

To evaluate f(x-1), we need to substitute x-1 into the function f(x) = 4x^2 + 8x - 5.

f(x-1) = 4(x-1)^2 + 8(x-1) - 5 f(x-1) = 4(x^2 - 2x + 1) + 8x - 8 - 5 f(x-1) = 4x^2 - 8x + 4 + 8x - 8 - 5 f(x-1) = 4x^2 - 9

Therefore, f(x-1) = 4x^2 - 9.

Conclusion

In this article, we have evaluated the given quadratic function f(x) = 4x^2 + 8x - 5 at specific values of x. We have also evaluated the function at x+1 and x-1. The results are as follows:

f(0) = -5
f(2) = 27
f(-2) = -5
f(x+1) = 4x^2 + 16x + 7
f(x-1) = 4x^2 - 9

Introduction

In our previous article, we explored the concept of evaluating a quadratic function at specific values of x. A quadratic function is a polynomial function of degree two, which means the highest power of the variable (in this case, x) is two. The general form of a quadratic function is f(x) = ax^2 + bx + c, where a, b, and c are constants.

Q&A

Q: What is a quadratic function?

A: A quadratic function is a polynomial function of degree two, which means the highest power of the variable (in this case, x) is two. The general form of a quadratic function is f(x) = ax^2 + bx + c, where a, b, and c are constants.

Q: How do I evaluate a quadratic function at a specific value of x?

A: To evaluate a quadratic function at a specific value of x, you need to substitute the value of x into the function. For example, if the function is f(x) = 4x^2 + 8x - 5 and you want to evaluate it at x = 2, you would substitute x = 2 into the function and simplify.

Q: What is the difference between f(x) and f(x+1)?

A: f(x) is the original function, while f(x+1) is the function evaluated at x+1. For example, if the function is f(x) = 4x^2 + 8x - 5, then f(x+1) = 4(x+1)^2 + 8(x+1) - 5.

Q: How do I evaluate f(x+1) and f(x-1)?

A: To evaluate f(x+1) and f(x-1), you need to substitute x+1 and x-1 into the function, respectively. For example, if the function is f(x) = 4x^2 + 8x - 5, then f(x+1) = 4(x+1)^2 + 8(x+1) - 5 and f(x-1) = 4(x-1)^2 + 8(x-1) - 5.

Q: What is the significance of evaluating a quadratic function at specific values of x?

A: Evaluating a quadratic function at specific values of x can help you understand the behavior of the function at those points. It can also help you identify the maximum or minimum value of the function, as well as the x-intercepts.

Q: How do I find the x-intercepts of a quadratic function?

A: To find the x-intercepts of a quadratic function, you need to set the function equal to zero and solve for x. For example, if the function is f(x) = 4x^2 + 8x - 5, you would set it equal to zero and solve for x: 4x^2 + 8x - 5 = 0.

Q: What is the relationship between the x-intercepts and the graph of a quadratic function?

A: The x-intercepts of a quadratic function are the points where the graph of the function intersects the x-axis. The graph of a quadratic function is a parabola, and the x-intercepts are the points where the parabola intersects the x-axis.

Conclusion

In this article, we have answered some common questions about evaluating a quadratic function at specific values of x. We have also discussed the significance of evaluating a quadratic function at specific values of x and how it can help you understand the behavior of the function at those points. We hope this article has provided a clear understanding of the concept of evaluating a quadratic function at specific values of x.