Problem 65 - Algebra and Trigonometry

Question

Sketch the function represented by the given parametric equations. Then use the graph to determine each of the following: a. intervals, if any, on which the function is increasing and intervals, if any, on which the function is decreasing. b. the number, if any, at which the function has a maximum and this maximum value, or the number, if any, at which the function has a minimum and this minimum value. \(x=\frac{t}{2}, y=2 t^{2}-8 t+3\)

Step-by-Step Solution

Answer

The function \(y=2x^{2}-8x+3\) is decreasing in the interval \(-∞, 2\) and increasing in the interval \(2, ∞\). The function has a minimum value of -5 at \(x=2\).

Step 1: Graphing the Function

The function which needs to be considered is \(y=2x^{2}-8x+3\). With the help of graphing, sketch the curve. Here \(x\) is parameterized by \(t\), but since \(x=t/2\), this doesn't change the shape of the graph of \(y\) as a function of \(x\).

Step 2: Calculating the Derivative

The derivative can be found by differentiating the function \(y=2x^{2}-8x+3\) with respect to \(x\). The derivative equation is \(y'=4x-8\).

Step 3: Finding Increasing and Decreasing Intervals

The function is increasing when the derivative \(y'\) is positive and it is decreasing when \(y'\) is negative. Setting \(y'=0\) gives \(x=2\). For \(x<2\), \(y'<0\) and the function is decreasing. For \(x>2\), \(y'>0\) and the function is increasing. Thus, the interval where the function is decreasing is \(-∞, 2\) and where it is increasing is \(2, ∞\).

Step 4: Finding Maximum or Minimum Values

For maximum and minimum values, calculate the values of \(y\) at the turning points which is at \(x=2\). Insert the value \(x=2\) into the equation \(y=2x^{2}-8x+3\). \(y(2)=-5\), thus the function has a minimum value of \(y=-5\) at \(x=2\).

Key Concepts

Increasing Intervals

In the context of parametric equations, understanding increasing intervals can be quite intuitive. Here's how it works: an interval is considered to be increasing if, when moving from left to right on the graph, the function's values consistently go up. In mathematical terms, this happens when the derivative of the function, denoted by \( y' \), is positive. With these parametric equations, the function becomes \( y = 2x^2 - 8x + 3 \).

To identify the increasing interval, we need to calculate the derivative. As provided, \( y' = 4x - 8 \). Setting this expression greater than zero helps find out when the function is increasing:

• \( 4x - 8 > 0 \)
• \( 4x > 8 \)
• \( x > 2 \)

This tells us that for \( x > 2 \), the values of \( y \) increase, meaning the function is on an increasing interval. Therefore, the increasing interval is \( (2, \infty) \).

Decreasing Intervals

A decreasing interval is where the function's values consistently drop as you move along the graph. For parametric equations like ours, a function is considered to be decreasing when its derivative \( y' \) is negative. In our case, the derivative is \( y' = 4x - 8 \).

To find when the function decreases, set the derivative less than zero:

• \( 4x - 8 < 0 \)
• \( 4x < 8 \)
• \( x < 2 \)

This indicates that for values of \( x \) less than 2, the function is decreasing. Thus, the interval \( (-\infty, 2) \) is where the function is decreasing. Understanding these intervals is crucial in analyzing the behavior of a function and predicting its path on a graph.

Derivative

The derivative is a fundamental tool in calculus that helps determine the rate at which a function is changing at any given point. It's essentially the function's slope at a specific point. In parametric equations, like the one we're working with, calculating the derivative \( y' \) is essential to understand the behavior of the function.
For the function \( y = 2x^2 - 8x + 3 \), the derivative is computed as \( y' = 4x - 8 \). This tells us how the function's values (or 'y-values') change as \( x \) changes.

• If \( y' > 0 \), the function is increasing.
• If \( y' < 0 \), the function is decreasing.
• If \( y' = 0 \), the function might have a local maximum or minimum (a turning point).

This makes derivatives incredibly powerful in sketching and predicting graphs. Determining where the derivative equals zero helps find potential turning points, which can further help identify maximum or minimum values of the function, making them a vital part of mathematical analysis.