Math Problem Statement

Sketching the Graph of f(x) = 3x^{3} - 4x^{2} - 5x + 2

Solution

To sketch the graph of the function f(x)=3x34x25x+2f(x) = 3x^3 - 4x^2 - 5x + 2, let's go through a step-by-step process.

1. Determine the Behavior of the Function (End Behavior):

  • Since this is a cubic polynomial (x3x^3 term is the highest power), the end behavior will follow the pattern of a cubic function.
  • The coefficient of x3x^3 is positive (3), so as xx \to \infty, f(x)f(x) \to \infty and as xx \to -\infty, f(x)f(x) \to -\infty.

2. Find the First Derivative (for Critical Points):

  • The first derivative, f(x)f'(x), will help identify any critical points (local maxima, minima).

  • f(x)=3x34x25x+2f(x) = 3x^3 - 4x^2 - 5x + 2

  • Differentiating with respect to xx: f(x)=9x28x5f'(x) = 9x^2 - 8x - 5

  • Set f(x)=0f'(x) = 0 to find critical points: 9x28x5=09x^2 - 8x - 5 = 0

  • Solving this quadratic equation using the quadratic formula: x=(8)±(8)249(5)29x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 9 \cdot (-5)}}{2 \cdot 9} x=8±64+18018x = \frac{8 \pm \sqrt{64 + 180}}{18} x=8±244188±15.6218x = \frac{8 \pm \sqrt{244}}{18} \approx \frac{8 \pm 15.62}{18} x1.3andx0.42x \approx 1.3 \quad \text{and} \quad x \approx -0.42

These are approximate values for the critical points.

3. Find the Second Derivative (for Concavity):

  • The second derivative, f(x)f''(x), helps determine the concavity of the function.

  • Differentiating f(x)f'(x): f(x)=18x8f''(x) = 18x - 8

  • Set f(x)=0f''(x) = 0 to find potential inflection points: 18x8=018x - 8 = 0 x=818=490.44x = \frac{8}{18} = \frac{4}{9} \approx 0.44

At x0.44x \approx 0.44, there is a potential inflection point where the concavity changes.

4. Evaluate Key Points (Including Intercepts):

  • Y-intercept: Set x=0x = 0: f(0)=3(0)34(0)25(0)+2=2f(0) = 3(0)^3 - 4(0)^2 - 5(0) + 2 = 2 So, the y-intercept is (0,2)(0, 2).

  • X-intercepts: Solve f(x)=0f(x) = 0: Solving 3x34x25x+2=03x^3 - 4x^2 - 5x + 2 = 0 may require numerical or graphical methods as it does not factor nicely.

5. Plot and Sketch the Graph:

Using the information gathered:

  • Plot the intercepts and mark critical points around x1.3x \approx 1.3 and x0.42x \approx -0.42.
  • Mark the inflection point around x0.44x \approx 0.44.
  • Show the end behavior: the graph moves up to infinity as xx \to \infty and down to negative infinity as xx \to -\infty.

If you'd like a visual graph plotted from these steps or have questions on any step, feel free to ask!


Follow-Up Questions:

  1. What are critical points, and how are they identified on a graph?
  2. How does concavity affect the shape of a function's graph?
  3. Why is the end behavior important in understanding polynomial functions?
  4. How can the quadratic formula help solve for critical points in a cubic function?
  5. What is the significance of an inflection point on a graph?

Tip:

To sketch polynomials accurately, start by finding intercepts, critical points, and concavity. These elements together guide the overall shape of the graph.

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Math Problem Analysis

Mathematical Concepts

Cubic Functions
Calculus
Derivatives
Critical Points
Inflection Points
End Behavior
Polynomial Graphs

Formulas

First Derivative: f'(x) = 9x^2 - 8x - 5
Quadratic Formula: x = [-b ± sqrt(b^2 - 4ac)] / (2a)
Second Derivative: f''(x) = 18x - 8

Theorems

Fundamental Theorem of Algebra
Derivative Test for Critical Points
Second Derivative Test for Concavity

Suitable Grade Level

Grades 11-12, College Level