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

Let's break it down:

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

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

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

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

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

Summary:

• The correct values of $x$ where the tangent is horizontal are $x = 1$ and $x = -3$.
• The corresponding points on the curve are $(1, 1)$ and $(-3, 33)$, 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 $x^2 + 2x - 3$?
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!