Given $h(t) = -2(t+5)^2 + 4$, Find $h(-8)$.A. { -334$}$ B. { -14$}$ C. 45 D. 445

Introduction

In mathematics, quadratic functions are a fundamental concept in algebra and calculus. They are used to model various real-world phenomena, such as the trajectory of a projectile, the growth of a population, and the motion of an object under the influence of a force. In this article, we will focus on evaluating a quadratic function at a specific point, using the given function $h(t) = -2(t+5)^2 + 4$ as an example.

Understanding the Function

The given function is a quadratic function in the form of $h(t) = a(t-h)^2 + k$, where $a$, $h$, and $k$ are constants. In this case, $a = -2$, $h = -5$, and $k = 4$. The function represents a parabola that opens downward, since $a$ is negative.

Evaluating the Function at a Specific Point

To evaluate the function at a specific point, we need to substitute the value of $t$ into the function. In this case, we want to find $h(-8)$. To do this, we will replace $t$ with $-8$ in the function.

Step 1: Substitute the Value of t

$$h(-8) = -2(-8+5)^2 + 4$$

Step 2: Simplify the Expression

$$h(-8) = -2(-3)^2 + 4$$

Step 3: Evaluate the Expression

$$h(-8) = -2(9) + 4$$

$$h(-8) = -18 + 4$$

$$h(-8) = -14$$

Conclusion

In this article, we evaluated the quadratic function $h(t) = -2(t+5)^2 + 4$ at the specific point $t = -8$. We followed a step-by-step approach to simplify the expression and arrive at the final answer. The correct answer is $h(-8) = -14$.

Discussion

This problem is a great example of how to evaluate a quadratic function at a specific point. It requires a good understanding of the function and its properties, as well as the ability to simplify expressions and evaluate them. The concept of quadratic functions is essential in mathematics and has many real-world applications.

Related Topics

• Quadratic functions
• Evaluating functions
• Simplifying expressions
• Algebra
• Calculus

Practice Problems

1. Evaluate the function $f(x) = 2(x-2)^2 + 1$ at the point $x = 4$.
2. Find the value of $g(3)$ for the function $g(x) = -3(x+2)^2 + 5$.
3. Evaluate the function $h(t) = t^2 - 4t + 3$ at the point $t = 2$.

Answer Key

1. $$f(4) = 2(2)^2 + 1 = 9$$

2. $$g(3) = -3(3+2)^2 + 5 = -45$$

3. h(2) = (2)^2 - 4(2) + 3 = -1$<br/>

Introduction

In our previous article, we explored the concept of evaluating quadratic functions at specific points. We used the function $h(t) = -2(t+5)^2 + 4$ as an example and walked through the steps to find $h(-8)$. In this article, we will continue to delve into the world of quadratic functions and answer some frequently asked questions.

Q&A

Q: What is a quadratic function?

A: A quadratic function is a polynomial function of degree two, which means it has the form $f(x) = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants.

Q: What is the general form of a quadratic function?

A: The general form of a quadratic function is $f(x) = a(x-h)^2 + k$, where $a$, $h$, and $k$ are constants.

Q: What is the vertex of a quadratic function?

A: The vertex of a quadratic function is the point on the graph where the function changes direction. It is represented by the coordinates $(h, k)$ in the general form of the function.

Q: How do I find the vertex of a quadratic function?

A: To find the vertex of a quadratic function, you need to find the values of $h$ and $k$ in the general form of the function. The vertex is then represented by the coordinates $(h, k)$.

Q: What is the axis of symmetry of a quadratic function?

A: The axis of symmetry of a quadratic function is a vertical line that passes through the vertex of the function. It is represented by the equation $x = h$.

Q: How do I find the axis of symmetry of a quadratic function?

A: To find the axis of symmetry of a quadratic function, you need to find the value of $h$ in the general form of the function. The axis of symmetry is then represented by the equation $x = h$.

Q: What is the x-intercept of a quadratic function?

A: The x-intercept of a quadratic function is the point on the graph where the function crosses the x-axis. It is represented by the coordinates $(x, 0)$.

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

A: To find the x-intercept of a quadratic function, you need to set the function equal to zero and solve for $x$.

Q: What is the y-intercept of a quadratic function?

A: The y-intercept of a quadratic function is the point on the graph where the function crosses the y-axis. It is represented by the coordinates $(0, y)$.

Q: How do I find the y-intercept of a quadratic function?

A: To find the y-intercept of a quadratic function, you need to substitute $x = 0$ into the function and solve for $y$.

Conclusion

In this article, we answered some frequently asked questions about quadratic functions. We covered topics such as the general form of a quadratic function, the vertex, axis of symmetry, x-intercept, and y-intercept. We hope that this article has provided you with a better understanding of quadratic functions and how to work with them.

Practice Problems

1. Find the vertex of the quadratic function $f(x) = 2(x-3)^2 + 1$.
2. Find the axis of symmetry of the quadratic function $g(x) = -3(x+2)^2 + 5$.
3. Find the x-intercept of the quadratic function $h(x) = x^2 - 4x + 3$.
4. Find the y-intercept of the quadratic function $f(x) = 2x^2 + 3x - 1$.

Answer Key

1. The vertex is $(3, 1)$.
2. The axis of symmetry is $x = -2$.
3. The x-intercept is $(1, 0)$.
4. The y-intercept is $(0, -1)$.

Related Topics

• Quadratic functions
• Evaluating functions
• Simplifying expressions
• Algebra
• Calculus

Discussion

Quadratic functions are a fundamental concept in mathematics and have many real-world applications. They are used to model various phenomena, such as the trajectory of a projectile, the growth of a population, and the motion of an object under the influence of a force. In this article, we covered some frequently asked questions about quadratic functions and provided practice problems to help you understand the concepts better.