To find the points on the curve $y = x^3 + 3x^2 - 9x + 6$ where the tangent is horizontal, we need to find the values of $x$ where the derivative of $y$ with respect to $x$ is zero. This is because the slope of the tangent line is given by the derivative, and a horizontal tangent means the slope is zero.

1. Find the derivative $y'$:

$y' = \frac{dy}{dx} = 3x^2 + 6x - 9$

2. Set the derivative equal to zero to find the critical points:

$3x^2 + 6x - 9 = 0$

3. Solve the quadratic equation:

$x^2 + 2x - 3 = 0$

Factoring:

$(x + 3)(x - 1) = 0$

So, the solutions are:

$x = -3 \quad \text{and} \quad x = 1$

4. Find the corresponding $y$-coordinates for each $x$-value:

For $x = -3$: $y = (-3)^3 + 3(-3)^2 - 9(-3) + 6 = -27 + 27 + 27 + 6 = 33$

So, the point is $(-3, 33)$.

For $x = 1$: $y = (1)^3 + 3(1)^2 - 9(1) + 6 = 1 + 3 - 9 + 6 = 1$

So, the point is $(1, 1)$.

5. Conclusion:

• Smaller $x$-value point: $(-3, 33)$
• Larger $x$-value point: $(1, 1)$

Would you like more details on this problem or have any questions?

Related questions:

1. What are the coordinates of points where the derivative is positive?
2. How do you find the points where the curve has a vertical tangent?
3. How do you determine if the critical points are maxima, minima, or saddle points?
4. Can you explain how to find inflection points on this curve?
5. What is the geometric meaning of the second derivative for this curve?

Tip: When solving a quadratic equation, always check if it can be factored easily before using the quadratic formula.