Problem 55 - College Algebra and Calculus: An Applied Approach

Question

Analyzing a Graph In Exercises \(47-58\), analyze the graph of the function algebraically and use the results to sketch the graph by hand. Then use a graphing utility to confirm your sketch. $$g(t)=-\frac{1}{4}(t-2)^{2}(t+2)^{2}$$

Step-by-Step Solution

Answer

The graph is an inverted quartic function with roots/zeros at \(t=-2\) and \(t=2\), a y-intercept at \(t=0, g = 4\), and even symmetry. It appears as a symmetric, inverted 'W' shape when sketched.

Step 1: Finding the zeroes/roots

To find the zeroes of the function, set \(g(t)\) equal to 0. Therefore, we have \(0=-\frac{1}{4}(t-2)^{2}(t+2)^{2}\). This tells us that the zeroes (where the function crosses the x-axis) are at \(t=-2\) and \(t=2\).

Step 2: Determine y-intercept

The y-intercept is the function value when \(t = 0\). Thus, place \(0\) into the function to get: \(g(0)=-\frac{1}{4}(2)^{2}(\-2)^{2} = 4\). This is the point where the graph crosses the y-axis, at (0,4).

Step 3: Identify Symmetry

The function \(g(t)=-\frac{1}{4}(t-2)^{2}(t+2)^{2}\) has even symmetry because replacing \(t\) with \(-t\) yields the same value. This means the graph will be symmetric about the y-axis.

Step 4: Identify local Minima or Maxima

The function \(g(t)=-\frac{1}{4}(t-2)^{2}(t+2)^{2}\) represents an inverted (due to the negative coefficient) quartic function with roots at \(-2\) and \(2\). Hence, it will have local maxima at these points.

Step 5: Finalize The Sketch and Use a Graphing Utility to Confirm

With the roots at \(t=-2\) and \(t=2\), maxima at these points as well, and the y-intercept at \(t=0, g = 4\), sketch the curve as a symmetric, inverted 'W' shape flipping about the y-axis. Use a graphing utility to ensure the sketch is correct.

Key Concepts

Zeroes of a Function

In the world of mathematics, finding the zeroes of a function is like discovering the points where a graph touches or crosses the x-axis. These points, often called roots, indicate where the function's value is zero. For the function \( g(t) = -\frac{1}{4}(t-2)^{2}(t+2)^{2} \), we discover the zeroes by setting the entire function equal to zero:

• \( 0 = -\frac{1}{4}(t-2)^{2}(t+2)^{2} \)
• This simplifies to finding \( t \) values that make the expression zero.
• The zeroes are found at \( t = -2 \) and \( t = 2 \).

These zeroes indicate where our graph will intersect the x-axis, showing us crucial turning points in the graph's behavior.

Y-intercept

The y-intercept is the point where the graph crosses the y-axis. To find this for any function, we simply evaluate the function at \( t = 0 \). For the given function \( g(t) = -\frac{1}{4}(t-2)^{2}(t+2)^{2} \), we substitute 0 for \( t \):

• \( g(0) = -\frac{1}{4}(2)^{2}(-2)^{2} \)
• This simplifies to \( g(0) = 4 \).

Thus, the y-intercept is at the point \( (0, 4) \). This is a crucial point in understanding how the graph interacts with the y-axis.

Symmetry

Symmetry in graphs can simplify our understanding by showing if one side of the graph mirrors the other. For the function \( g(t) = -\frac{1}{4}(t-2)^{2}(t+2)^{2} \), this symmetry is about the y-axis. This conclusion comes because substituting \( -t \) for \( t \) results in the same expression:

• \( g(-t) = -\frac{1}{4}(-t-2)^{2}(-t+2)^{2} \)
• Simplifies to \( g(t) = -\frac{1}{4}(t-2)^{2}(t+2)^{2} \)

This form confirms that the function is even, meaning it is symmetric about the y-axis. Recognizing symmetry helps when sketching the function because it reflects the functional behavior on one side to the other.

Local Minima and Maxima

Finding local minima and maxima helps to identify the highest and lowest points in certain intervals of a graph. For our function \( g(t) = -\frac{1}{4}(t-2)^{2}(t+2)^{2} \), we can expect certain crucial points. Due to the negative coefficient, this quartic function opens downward, resembling an inverted 'W' shape. The crucial aspects include:

• Local maxima appear at the roots: \( t = -2 \) and \( t = 2 \).
• These points reflect the peaks of the curve, indicating the turning points where the function changes direction from increasing to decreasing.

Understanding these maxima aids in constructing an accurate sketch of the function’s behavior across its domain.