Problem 97 - College Algebra

Question

Use long division to rewrite the equation for \(g\) in the form $$\text {quotient }+\frac{\text {remainder}}{\text {divisor}}$$ Then use this form of the function's equation and transformations. $$g(x)=\frac{3 x-7}{x-2}$$

Step-by-Step Solution

Answer

The function \(g(x)\) rewritten in the form of quotient + remainder/divisor is \(g(x) = 3 + \frac{6}{x - 2}\)

Step 1: Performing Long Division

Set up the long division with the numerator as the dividend and the denominator as the divisor. \\[\frac{3x - 7}{x - 2}\\] This is similar to arithmetic long division. Divide the first term of the dividend (3x) by the first term of the divisor (x).

Step 2: Multiply and Subtract

The result from step 1 is 3 (the quotient). Multiply the entire divisor by 3, subtract it from the dividend and bring down the next term. \\[3x - (3 * (x - 2)) \\] This simplifies to \\[3x - 3x + 6 = 6\\]

Step 3: Write the function in the form of quotient + remainder/divisor

Adding the remainder to the product of quotient and divisor gives the original dividend. So, \(g(x)\) can be rewritten as \\[g(x) = 3 + \frac{6}{x - 2}\\]

Key Concepts

Long Division in Rational Functions

Long division is a crucial concept when it comes to breaking down complex rational functions into simpler components. A rational function is essentially a fraction where both the numerator and the denominator are polynomials. To simplify such functions, long division can be applied, much like with numbers. Here's how you should do it:

• Start with the numerator as the dividend and the denominator as the divisor. In our example, the dividend is \(3x - 7\) and the divisor is \(x - 2\).
• Divide the first term of the dividend by the first term of the divisor. Here, that's \(\frac{3x}{x}\), which equals 3. This is the beginning of your quotient.
• Next, multiply the entire divisor (\(x - 2\)) by this quotient term (3), which gives \(3(x - 2)\).
• Subtract this product from the original dividend to find the new dividend. In this instance, \(3x - (3x - 6) = 6\).

This step-by-step method reveals that our function can be rewritten in a form that is easier to analyze, namely as the quotient plus the remainder over the original divisor.

Understanding Transformations in Functions

Transformations are like the moves that change the shape or position of a graph on the coordinate plane. When you apply the concept of transformations to a rational function rewritten from our long division, you gain insights into how the graph behaves and where it sits. The function \(g(x) = 3 + \frac{6}{x-2}\) is derived from the division:

• The constant 3 is a vertical shift, moving the graph up by three units.
• The fraction \(\frac{6}{x-2}\) brings in a horizontal component. The denominator's \(x-2\) implies a horizontal shift rightward by two units, as it affects the \(x\) component directly.

Transformations allow us to visually understand these changes:

• Vertical shifts happen when you add or subtract a constant from the entire function.
• Horizontal shifts occur when constants are involved inside the function's argument, like \(x-2\).

These shifts are part of the structural transformations that reshape how we interpret the behavior of rational functions on a graph.

Role of Remainders in Function Representation

When you perform long division on polynomials to simplify a rational function, you often encounter a remainder. But what exactly is a remainder and what does it mean in the context of rational functions? The remainder is what is left over after dividing the polynomials completely.In our example, where the original function \(g(x) = \frac{3x - 7}{x - 2}\) is rewritten as \(g(x) = 3 + \frac{6}{x - 2}\), the number 6 represents the remainder:

• The "6" is linked to what's left after "3 times (x - 2)" is subtracted from \(3x - 7\).
• It is crucial because in the rewritten form of the function, this remainder over the divisor \(x-2\) significantly impacts the behavior of the function as \(x\) approaches certain values.

Remainders are not just leftovers; they play a vital part in expressing the function as a combination of simple components. They help us understand asymptotic behavior and guide us in predicting how the function behaves outside standard bounds.