Given The Function $f(x)=4x^2+8x-5$, Calculate The Following Values:1. F ( 0 ) = F(0)= F ( 0 ) = $\square$2. F ( 2 ) = F(2)= F ( 2 ) = $\square$3. F ( − 2 ) = F(-2)= F ( − 2 ) = $\square$4. F ( X + 1 ) = F(x+1)= F ( X + 1 ) = $\square$5.
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.