Question

The following transformations 𝑦 = 3𝑓 (1/2 𝑥 − 2𝜋) − 1 were applied to the parent function f(x) =csc(x). Graph the transformed function for the interval −4𝜋 ≤ 𝑥 ≤ 4𝜋. On your graph, label the

asymptotes, local max/min, and number each axis. Use mapping notation to show your work for a
minimum of 5 key points for full marks. [6C]

Answer 1

[AI-generated answer — Explain Bot]

To graph the transformed function, we need to apply each transformation step by step on the parent function f(x) = csc(x).

The given transformations are:
y = 3f(1/2x - 2π) - 1

Step 1: Starting with the parent function f(x) = csc(x), let's plot the graph of the parent function over the given interval −4π ≤ x ≤ 4π.

To map the key points, we need to determine the values of x that correspond to certain key points on the graph.

For the parent function f(x) = csc(x), the asymptotes occur at x = 0, ±π, ±2π, ±3π, ....

Now, let's focus on the interval −4π ≤ x ≤ 4π and find the values of x for the selected points of interest.

1. Local Maximum:
The first local maximum occurs at x = π/2.

2. Local Minimum:
The first local minimum occurs at x = -π/2.

3. Increasing Interval:
Between the local minimum and the local maximum, we can choose a point.

4. Decreasing Interval:
Between the local maximum and the local minimum, we can choose a point.

5. A point on an asymptote:
Let's choose a point near the asymptote x = π.

Now that we have selected the key points, we can plug these values of x into the given transformation to find their corresponding y values.

Step 2: Apply the transformations to the parent function:
For each key point, we will apply the transformations one by one to determine the corresponding y values.

1. Local Maximum:
x = π/2
First, apply the transformation f(1/2x - 2π) - 1:
f(1/2(π/2) - 2π) - 1 = f(-3π/4) - 1

Find the corresponding y value by evaluating the parent function at the transformed x value:
f(-3π/4) - 1 = csc(-3π/4) - 1

2. Local Minimum:
x = -π/2
Apply the transformations:
f(1/2x - 2π) - 1 = f(1/2(-π/2) - 2π) - 1 = f(-5π/4) - 1

Find the corresponding y value:
f(-5π/4) - 1 = csc(-5π/4) - 1

3. Increasing Interval:
Choose a point such as x = −π/4.

Apply the transformations:
f(1/2x - 2π) - 1 = f(1/2(-π/4) - 2π) - 1 = f(-9π/8) - 1

Find the corresponding y value:
f(-9π/8) - 1 = csc(-9π/8) - 1

4. Decreasing Interval:
Choose a point such as x = 3π/4.

Apply the transformations:
f(1/2x - 2π) - 1 = f(1/2(3π/4) - 2π) - 1 = f(-π/4) - 1

Find the corresponding y value:
f(-π/4) - 1 = csc(-π/4) - 1

5. A point on an asymptote:
Choose a point near x = π.

Apply the transformations:
f(1/2x - 2π) - 1 = f(1/2(π) - 2π) - 1 = f(-3π/2) - 1

Find the corresponding y value:
f(-3π/2) - 1 = csc(-3π/2) - 1

Now that we have determined the corresponding y values for the key points, we can plot them on the graph. We will label the asymptotes, local maxima and minima, and number each axis accordingly.