Problem 33 - Algebra and Trigonometry Real Mathematics, Real People

Question

Evaluate the function at each specified value of the independent variable and simplify. $$h(t)=t^{2}-2 t$$ (a) \(h(2)\) (b) \(h(1.5)\) (c) \(h(x-4)\)

Step-by-Step Solution

Answer

The evaluated function for the given points: (a) \(h(2) = 2\), (b) \(h(1.5) = -0.75\), (c) \(h(x-4) = x^2 - 10x + 20\)

Step 1: Substituting t with 2

Replace \(t\) with 2 in the equation \(h(t) = t^2 - 2t\). So, \(h(2) = (2)^2 - 2*(2)\)

Step 2: Simplify the equation for h(2)

Calculate \(h(2)\). We get \( h(2) = 2\)

Step 3: Substituting t with 1.5

Replace \(t\) with 1.5 in the equation \(h(t) = t^2 - 2t\). So, \(h(1.5) = (1.5)^2 - 2*(1.5)\)

Step 4: Simplify the equation for h(1.5)

Calculate \(h(1.5)\). We get \( h(1.5) = -0.75\)

Step 5: Substituting t with x-4

Replace \(t\) with \(x - 4\) in the equation \(h(t) = t^2 - 2t\). So, \(h(x-4) = (x-4)^2 - 2*(x-4)\)

Step 6: Simplify the equation for h(x-4)

Expand the equation for \(h(x-4)\) to obtain: \(h(x-4) = x^2 - 10x + 20\)

Key Concepts

Polynomial Functions

Polynomial functions are a type of mathematical expression involving a sum of powers of a variable. Each power is associated with a coefficient, and these functions are frequently used to model various real-life phenomena. In the case of the function \( h(t) = t^2 - 2t \), it is a polynomial of degree two. This is because the variable \( t \) is raised to the second power as the highest degree.
Polynomials are classified according to their degrees:

• Constant polynomial: degree 0 (e.g., \( 5 \))
• Linear polynomial: degree 1 (e.g., \( 3t + 2 \))
• Quadratic polynomial: degree 2 (e.g., \( h(t) = t^2 - 2t \))
• Cubic polynomial: degree 3 (e.g., \( t^3 - 3t^2 + t + 1 \))

Understanding the degree of a polynomial is crucial, as it directly influences the shape of its graph and the behavior of the function. For quadratic polynomials like our example, the graph is a parabola, which can open upwards or downwards based on the sign of the leading coefficient (here, it's positive, so the parabola opens upwards).

Substitution Method

The substitution method in mathematics involves replacing a variable in an equation or expression with a given value. This approach simplifies evaluating functions, especially when specific values for the variables are known.
For the function \( h(t) = t^2 - 2t \), the problem asked us to evaluate it at particular values: \( t = 2 \), \( t = 1.5 \), and \( t = x - 4 \). Here’s how substitution works:

• Replace \( t \) with \( 2 \) to find \( h(2) \).
• Replace \( t \) with \( 1.5 \) to get \( h(1.5) \).
• Replace \( t \) with \( x-4 \) for \( h(x-4) \).

This method allows one to compute the value of the function at multiple points effortlessly, which is particularly useful in applications like graphing or solving for specific conditions.Using the substitution method helps demystify the process of finding function values and ensures that solutions are obtained using a systematic approach. It essentially allows us to "plug and play" with different values, revealing how the function behaves under various scenarios.

Simplification

Simplification is the process of reducing a mathematical expression to its simplest form. This step is crucial after substitution, as it converts a complex expression into a more manageable form.
Throughout the problem, simplification appears after each substitution to achieve a clearer solution:

• For \( h(2) \), we have \( (2)^2 - 2 \times 2 = 4 - 4 \) leading to \( h(2) = 0 \).
• For \( h(1.5) \), compute \( (1.5)^2 - 2 \times 1.5 = 2.25 - 3 \), which simplifies to \( h(1.5) = -0.75 \).
• For \( h(x-4) \), the expansion is \( (x-4)^2 - 2(x-4) = x^2 - 8x + 16 - 2x + 8 \), which simplifies to \( x^2 - 10x + 20 \).

Simplification not only helps in obtaining the correct answer but also makes these answers easier to understand and interpret. Additionally, it is an essential skill in algebra, as simplified expressions are often easier to compare, manipulate, or further analyze in more complex calculations.