The task asks to sketch the graph of the function $f(x) = x^3 - 5$. Here's a step-by-step explanation to help sketch this graph:

Step 1: Understand the Function

The function $f(x) = x^3 - 5$ is a cubic function. Its general shape will resemble an $x^3$ curve but shifted vertically down by 5 units due to the $-5$.

Step 2: Key Features

1. Intercepts:

   • Y-intercept: Set $x = 0$: $f(0) = 0^3 - 5 = -5.$ The y-intercept is $(0, -5)$.
   • X-intercept: Set $f(x) = 0$: $x^3 - 5 = 0 \quad \Rightarrow \quad x^3 = 5 \quad \Rightarrow \quad x = \sqrt[3]{5}.$ Approximate $x \approx 1.71$.

2. End Behavior:

   • As $x \to \infty$, $f(x) \to \infty$.
   • As $x \to -\infty$, $f(x) \to -\infty$.

3. Symmetry:

   • The function is not symmetric because it is neither even nor odd.

Step 3: Plot Points

Calculate several points to better sketch the curve:

• $f(-2) = (-2)^3 - 5 = -8 - 5 = -13$.
• $f(-1) = (-1)^3 - 5 = -6$.
• $f(0) = -5$.
• $f(1) = 1^3 - 5 = -4$.
• $f(2) = 2^3 - 5 = 8 - 5 = 3$.

Step 4: Sketch the Graph

1. Plot the calculated points $(-2, -13), (-1, -6), (0, -5), (1, -4), (2, 3)$.
2. Draw a smooth curve passing through these points, keeping the end behavior and cubic shape in mind.

Would you like me to provide a detailed plot or clarify any of the steps?

Questions for Practice:

1. How does the graph of $f(x) = x^3 - 5$ differ from $f(x) = x^3$?
2. Find the slope of the tangent line to the curve at $x = 2$.
3. Determine whether $f(x) = x^3 - 5$ has any points of inflection.
4. What is the derivative of $f(x)$, and how does it help in analyzing the curve?
5. Sketch $f(x) = x^3 + 5$ and compare it with $f(x) = x^3 - 5$.

Tip:

For polynomial functions, always start by analyzing their intercepts, end behavior, and turning points to understand the overall graph shape.