Problem 29 - College Algebra with Corequisite Support

Question

For the following exercises, find the average rate of change of each function on the interval specified. \(h(x)=5-2 x^{2}\) on [-2,4]

Step-by-Step Solution

Answer

The average rate of change is -4.

Step 1: Understand the Average Rate of Change Formula

The average rate of change of a function over an interval \([a, b]\) is calculated using the formula:\[\frac{f(b) - f(a)}{b - a}\]Here, you are finding how much the function changes, on average, per unit increase in \(x\).

Step 2: Identify the Function and Interval

We are given the function \(h(x) = 5 - 2x^{2}\) and the interval is \([-2, 4]\). Therefore, \(a = -2\) and \(b = 4\).

Step 3: Calculate h(a) and h(b)

Calculate \(h(a)\) when \(a = -2\):\[h(-2) = 5 - 2(-2)^{2} = 5 - 2(4) = 5 - 8 = -3\]Calculate \(h(b)\) when \(b = 4\):\[h(4) = 5 - 2(4)^{2} = 5 - 2(16) = 5 - 32 = -27\]

Step 4: Substitute in the Average Rate of Change Formula

Substitute the values of \(h(a)\) and \(h(b)\) into the average rate of change formula:\[\frac{h(4) - h(-2)}{4 - (-2)} = \frac{-27 - (-3)}{4 + 2}\]Simplify the expression:\[\frac{-27 + 3}{6} = \frac{-24}{6} = -4\]

Step 5: Conclusion: State the Result

The average rate of change of \(h(x) = 5 - 2x^{2}\) on the interval \([-2, 4]\) is \(-4\). This indicates that, on average, the function decreases by 4 units for each unit increase in \(x\) across the interval.

Key Concepts

Quadratic Function

A quadratic function is a type of polynomial function that can be expressed in the general form \( f(x) = ax^2 + bx + c \), where \( a, b, \) and \( c \) are constants, and \( a eq 0 \). In essence, it describes a parabola in a coordinate plane. The graph of a quadratic function is a smooth, symmetrical U-shaped curve, which can open upwards or downwards depending on the sign of the coefficient \( a \).
Here, we have a specific quadratic function \( h(x) = 5 - 2x^2 \). The term \(-2x^2\) tells us that the parabola opens downwards because \( a = -2 \), which is less than zero. The term \(5\) is the constant, representing the y-intercept of the parabola, the point where the graph intersects the y-axis.
Quadratic functions play a crucial role in various fields, such as physics for modeling trajectories and in finance for risk analysis scenarios.

Interval Notation

Interval notation is a way of representing a portion of the number line. It specifies the start and end points of an interval. This notation can indicate whether endpoints are included or excluded by using brackets or parentheses.
- Square brackets \([ ]\) indicate that the endpoint is included in the interval (closed interval).- Parentheses \(( )\) imply that the endpoint is not included (open interval).
In the problem at hand, we are given the interval \([-2, 4]\). This means the interval includes both \(-2\) and \(4\). Here, all values between \(-2\) and \(4\), limits inclusive, are within our range. Interval notation is an efficient way to express these values, avoiding lengthy descriptions.
When working with functions and their respective domains or ranges, interval notation becomes invaluable for succinctly conveying possible values.

Function Evaluation

Function evaluation is the process of finding the output of a function for specific inputs. It involves substituting a value (or values) into a function and computing the result. This is an essential skill in algebra as it helps determine specific points on the function’s graph.

The problem involves the function \(h(x) = 5 - 2x^2\). By evaluating this function at the endpoints of the given interval, \(-2\) and \(4\), we can determine how the function behaves over that stretch.

- For \(a = -2\), we substitute into \(h(x)\): - \(h(-2) = 5 - 2(-2)^2 = 5 - 8 = -3\)
- For \(b = 4\), we substitute: - \(h(4) = 5 - 2(4)^2 = 5 - 32 = -27\)
Function evaluation allows us to calculate these values accurately, showing how the function changes from one point to another.

Algebraic Expressions

Algebraic expressions are mathematical phrases that can include numbers, variables, and operation symbols. These expressions do not have an equality sign, thus they do not form complete equations, but they are still central in solving algebraic problems.
In our problem, the quadratic function \( h(x) = 5 - 2x^2 \) is an algebraic expression. It combines constants (\(5\)), variables (\(x\)), and operators (subtraction and multiplication). Solving problems involving algebraic expressions often involves basic arithmetic operations and understanding how variables are manipulated.
The expression \(5 - 2x^2\) demonstrates simplification. The subtraction of \(2x^2\) from \(5\) must be carefully evaluated based on specific values of \(x\). This manipulation is crucial for evaluating the function over specified intervals and finding average rates of change.
Grasping how to work with algebraic expressions, therefore, aids in solving a wide range of mathematical problems, from basic arithmetic operations to more complex functions.