Math Problem Statement

Find the points where the graph of the function y(x) = x^3 + 3x^2 - 9x + 6 has a horizontal tangent line.

Solution

From the image, you're solving a calculus problem involving a function y(x)=x3+3x29x+6y(x) = x^3 + 3x^2 - 9x + 6, and you are working through finding the derivative y(x)y'(x), determining where the tangent is horizontal, and calculating the corresponding points.

Let's break it down:

  1. Find the derivative y(x)y'(x): Given the function: y(x)=x3+3x29x+6y(x) = x^3 + 3x^2 - 9x + 6 The derivative is: y(x)=3x2+6x9y'(x) = 3x^2 + 6x - 9 You correctly found the derivative, as shown in the image.

  2. Find where the tangent line is horizontal: A horizontal tangent line occurs when y(x)=0y'(x) = 0. So, you need to solve: 3x2+6x9=03x^2 + 6x - 9 = 0 Dividing the entire equation by 3 simplifies this to: x2+2x3=0x^2 + 2x - 3 = 0 Factor the quadratic: (x1)(x+3)=0(x - 1)(x + 3) = 0 So, x=1x = 1 or x=3x = -3.

  3. Find the points on the curve: Now, plug these values of xx back into the original function y(x)y(x) to find the corresponding yy-coordinates.

    • For x=1x = 1: y(1)=13+3(1)29(1)+6=1+39+6=1y(1) = 1^3 + 3(1)^2 - 9(1) + 6 = 1 + 3 - 9 + 6 = 1 So, one point is (1,1)(1, 1).

    • For x=3x = -3: y(3)=(3)3+3(3)29(3)+6=27+27+27+6=33y(-3) = (-3)^3 + 3(-3)^2 - 9(-3) + 6 = -27 + 27 + 27 + 6 = 33 So, the other point is (3,33)(-3, 33).

Summary:

  • The correct values of xx where the tangent is horizontal are x=1x = 1 and x=3x = -3.
  • The corresponding points on the curve are (1,1)(1, 1) and (3,33)(-3, 33), which are correct as indicated in the image.

Would you like further details on solving these types of problems or another question breakdown?

Here are five related questions:

  1. How do you determine if a function has a horizontal tangent line?
  2. What are the steps to factor a quadratic equation like x2+2x3x^2 + 2x - 3?
  3. What are the implications of the second derivative at the points of horizontal tangency?
  4. How do you verify if the points found are maxima, minima, or points of inflection?
  5. What changes in the process if the function includes trigonometric terms?

Tip: Always check if a quadratic equation can be factored before using the quadratic formula—it can save you time!

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Quadratic Equations

Formulas

y'(x) = 3x^2 + 6x - 9
Horizontal tangency occurs when y'(x) = 0
Quadratic equation: x^2 + 2x - 3 = 0
Factor: (x - 1)(x + 3) = 0

Theorems

Derivative test for horizontal tangents
Quadratic factoring

Suitable Grade Level

Grade 12 and introductory college-level Calculus