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I can see the image of a calculus midterm question about implicit differentiation and tangents. Let's go over each part step by step:

### Problem 4: Tangents and Implicit Differentiation
You are given the equation:
\[
x^2 - 3x = y^2 - 4y + 2
\]

#### a) **Implicit Differentiation:**
You need to find the derivative \( \frac{dy}{dx} \) using implicit differentiation.

- Differentiate both sides with respect to \( x \).
  \[
  \frac{d}{dx}(x^2 - 3x) = \frac{d}{dx}(y^2 - 4y + 2)
  \]
  \[
  2x - 3 = 2y \frac{dy}{dx} - 4 \frac{dy}{dx}
  \]
- Now, isolate \( \frac{dy}{dx} \) (denoted as \( y' \)):
  \[
  2x - 3 = y'(2y - 4)
  \]
  \[
  y' = \frac{2x - 3}{2y - 4}
  \]
  
This is the formula for the derivative \( y' \).

#### b) **Horizontal Tangents:**
For horizontal tangents, the slope \( y' = 0 \). Therefore, set the numerator of the derivative \( 2x - 3 = 0 \) and solve for \( x \):

\[
2x - 3 = 0 \implies x = \frac{3}{2}
\]

Substitute \( x = \frac{3}{2} \) into the original equation to find corresponding \( y \)-values.

#### c) **Vertical Tangents:**
For vertical tangents, the denominator \( 2y - 4 = 0 \). Solve for \( y \):

\[
2y - 4 = 0 \implies y = 2
\]

Substitute \( y = 2 \) into the original equation to find corresponding \( x \)-values.

#### d) **Equation of the Tangent Line at Point (2, 4):**
Use the point-slope form of the line equation:
\[
y - y_1 = m(x - x_1)
\]
First, find the slope \( m \) by substituting \( (x, y) = (2, 4) \) into \( y' \):

\[
y' = \frac{2(2) - 3}{2(4) - 4} = \frac{4 - 3}{8 - 4} = \frac{1}{4}
\]

Now, use the point \( (2, 4) \) and slope \( \frac{1}{4} \):
\[
y - 4 = \frac{1}{4}(x - 2)
\]
This simplifies to:
\[
y = \frac{1}{4}x + \frac{7}{2}
\]

---

Let me know if you would like more detailed steps or clarifications on any part.

Here are 5 related questions to practice:
1. How do you find the second derivative of an implicitly defined function?
2. What are the conditions for a curve to have a slant asymptote?
3. What happens to the tangent line when \( \frac{dy}{dx} \) is undefined?
4. How do you find points of inflection for an implicitly defined curve?
5. Can implicit differentiation be applied to functions of more than two variables?

**Tip:** Always check if the denominator of the derivative equals zero to identify vertical tangents.

Find the implicit derivative of x^2 - 3x = y^2 - 4y + 2, determine points of horizontal and vertical tangents, and find the equation of the tangent line at a specific point.

Solve the calculus problem involving implicit differentiation for the equation x^2 - 3x = y^2 - 4y + 2. Learn how to find horizontal and vertical tangents and the equation of the tangent line at the point (2,4). Suitable for college-level calculus students.

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