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To solve this problem, we are asked to find the points on the graph of the function \( f(x) = 2x^2 - 3x + 3 \) where the tangent line is horizontal. A tangent line is horizontal when its slope is zero. The slope of the tangent line is given by the derivative of the function \( f(x) \).

### Steps:

1. **Find the derivative of \( f(x) \):**

   The derivative of the function \( f(x) = 2x^2 - 3x + 3 \) is:
   \[
   f'(x) = \frac{d}{dx}(2x^2 - 3x + 3)
   \]
   Using basic differentiation rules:
   \[
   f'(x) = 4x - 3
   \]

2. **Set the derivative equal to 0:**

   To find where the tangent line is horizontal, we set the derivative equal to zero and solve for \( x \):
   \[
   4x - 3 = 0
   \]
   Solving for \( x \):
   \[
   4x = 3
   \]
   \[
   x = \frac{3}{4}
   \]

3. **Find the corresponding \( y \)-coordinate:**

   To find the point where the tangent is horizontal, substitute \( x = \frac{3}{4} \) into the original function:
   \[
   f\left(\frac{3}{4}\right) = 2\left(\frac{3}{4}\right)^2 - 3\left(\frac{3}{4}\right) + 3
   \]
   Simplifying:
   \[
   f\left(\frac{3}{4}\right) = 2\left(\frac{9}{16}\right) - \frac{9}{4} + 3
   \]
   \[
   f\left(\frac{3}{4}\right) = \frac{18}{16} - \frac{9}{4} + 3 = \frac{9}{8} - \frac{18}{8} + \frac{24}{8}
   \]
   \[
   f\left(\frac{3}{4}\right) = \frac{15}{8}
   \]

Thus, the point where the tangent line is horizontal is \( \left( \frac{3}{4}, \frac{15}{8} \right) \).

### Final Answer:
The point where the tangent line is horizontal is \( \left( \frac{3}{4}, \frac{15}{8} \right) \).

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Would you like more details on any part of the solution or have any questions?

Here are 5 related questions to consider:
1. How do you find the derivative of higher-order polynomials?
2. What is the geometric interpretation of a horizontal tangent line?
3. How does the shape of the graph change near points where the tangent is horizontal?
4. What if a function has multiple points where the tangent is horizontal?
5. How do you determine if a critical point is a maximum, minimum, or neither?

**Tip:** The derivative represents the slope of the tangent line at any point on a function, and setting it equal to zero can help find important points like maxima, minima, or points of inflection.

For the function, find the points on the graph at which the tangent line is horizontal. If none exist, state that fact. The function is f(x) = 2x^2 - 3x + 3.

Learn how to find the points where the tangent line to the graph of f(x) = 2x^2 - 3x + 3 is horizontal. This problem covers key calculus concepts, including differentiation and solving for critical points. Suitable for students in Grades 11-12.

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