Question

describe the error in graphing (f(x)) for the function, then graph it correctly.

(rough graph only)!

(f(x) = (x - 4)(x + 1)^2)

Explanation:

Step1: Identify the correct x-intercepts

$$x - 4 = 0 \implies x = 4 \quad \text{and} \quad x + 1 = 0 \implies x = -1$$

Step2: Determine the multiplicity behavior

$$\text{At } x = -1 \text{ (multiplicity 2): tangent; at } x = 4 \text{ (multiplicity 1): crosses}$$

Step3: Find the y-intercept

$$f(0) = (0 - 4)(0 + 1)^2 = -4$$

Step4: Analyze end behavior

$$\lim_{x \to \infty} f(x) = \infty \quad \text{and} \quad \lim_{x \to -\infty} f(x) = -\infty$$

Answer:

The error in the graph is that the x-intercepts are plotted incorrectly. The graph shows a bounce at \(x = 1\) and a cross at \(x = -4\), which corresponds to the function \(g(x) = (x + 4)(x - 1)^2\). For the given function \(f(x) = (x - 4)(x + 1)^2\), the graph should have a bounce (tangency) at \(x = -1\) and cross the x-axis at \(x = 4\).

To graph it correctly:

1. Plot the x-intercepts at \(x = -1\) (where the graph is tangent to the x-axis) and \(x = 4\) (where the graph crosses the x-axis).
2. Plot the y-intercept at \((0, -4)\).
3. Sketch a cubic curve that starts in the third quadrant, goes up to touch the x-axis at \((-1, 0)\), turns down to pass through \((0, -4)\), reaches a local minimum, and then goes up to cross the x-axis at \((4, 0)\) into the first quadrant.