Problem 66 - 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. \(x=\frac{t}{2}, y=-2 t^{2}+8 t-1\)

Step-by-Step Solution

Answer

The function \(y = -8x^2 + 16x - 1\) is increasing on the interval (-\infty, 1) and decreasing on the interval (1, \infty). It has a maximum value of 7 at x = 1.

Step 1: Convert parametric equations into a single equation

Replace \(t\) in the equation \(y = -2t^2 + 8t - 1\) with \(2x\) from the equation \(x = t/2\) which gives us a new equation \(y = -2(2x)^2 + 8(2x) - 1 = -8x^2 + 16x - 1\). This is the equation of a downward-opening parabola and can be used to graph the function.

Step 2: Determine intervals of increase or decrease

From the equation \(y = -8x^2 + 16x - 1\) we can see that the function increases for x < 1 and decreases for x > 1. So the function is increasing on the interval \(x = (-\infty, 1)\) and decreasing on the interval \(x = (1, \infty)\). This can also be seen on the graph by observing where the slope is positive or negative.

Step 3: Determine Maximum and Minimum Values

Using the first derivative test, we can determine the maximum or minimum values of the equation. The first derivative of the equation \(y = -8x^2 + 16x - 1\) is \(y' = -16x + 16\). Setting \(y'\) equal to zero gives us \(x = 1\) as the only critical point. At this x-value, \(y = 7\) which is a maximum as the function is increasing before x = 1 and decreasing after. Thus, the function reaches its maximum of 7 at x = 1.

Key Concepts

Parabola

In mathematics, a parabola is a smooth, u-shaped curve that can open upwards or downwards. Parabolas come from quadratic equations, which are typically written in the form \(y = ax^2 + bx + c\). These are two-dimensional shapes but are often plotted on a graph to show how one quantity changes with another. The important feature of a parabola is its vertex, which is the highest or lowest point on the graph. The position of the vertex determines whether the parabola opens upward or downward:

• If the coefficient \(a\) is positive, the parabola opens upwards like a cup.
• If the coefficient \(a\) is negative, it opens downwards like an upside-down cup.

In the given exercise, the rewritten equation is \(y = -8x^2 + 16x - 1\). Here, \(a = -8\), indicating that the parabola opens downwards. The vertex, serving as the maximum point, can be found using this equation. Parabolas are important not just in mathematics, but also in physics and engineering, where they describe paths of projectiles and arches in structures.

First Derivative Test

The first derivative test is a method used to find the local maxima and minima of a function. In simple terms, it helps us identify whether a function is increasing or decreasing in certain intervals.To apply the first derivative test:

• First, find the derivative of the function, which gives us a new equation. This derivative represents the slope of the tangent line to the curve at any point.
• Next, set the derivative equal to zero and solve for \(x\). This will provide the critical points where the slope changes from positive to negative or vice versa.
• These critical points are then tested to determine if they correspond to a maximum, minimum, or neither.

In our case, the derivative of \(y = -8x^2 + 16x - 1\) is \(y' = -16x + 16\). Solving \(y'=0\) gives \(x=1\), which is our critical point. Using this test, we find the function increases until \(x=1\) and decreases afterwards, confirming \(x=1\) is a maximum point.

Critical Points

Critical points are vital in the analysis of graphs of functions. They are the \(x\)-values where the first derivative of a function is zero or undefined. These points are significant because they indicate potential locations for local maxima and minima.To identify critical points:

• First, take the derivative of the function.
• Then, solve for \(x\) by setting the derivative equal to zero. This will provide the critical points where changes occur in the function's increasing or decreasing nature.
• Analyze the behavior of the function around these points using tests like the first derivative test.

In our exercise, the derivative \(y' = -16x + 16\) provided a critical point at \(x = 1\). By analyzing the function's behavior around this point, we see it is a maximum, because the function shifts from increasing to decreasing at \(x = 1\). This is a clear indication that the critical point leads us to the maximum value of the function.