https://file.aiquickdraw.com/mathbot/file/1725526939767lx69o7az.jpg

Let's break down and solve the problems shown in the image one by one:

### 1. **Find the slope of the equation \( y = 2 - x^2 \) at the point \( (3, -7) \)**

The slope of a curve is given by the derivative of the function. So, we need to find \( \frac{dy}{dx} \) for the equation \( y = 2 - x^2 \), and then evaluate it at \( x = 3 \).

- Derivative of \( y = 2 - x^2 \):
  \[
  \frac{dy}{dx} = -2x
  \]
- At \( x = 3 \):
  \[
  \frac{dy}{dx} = -2(3) = -6
  \]
So, the slope at the point \( (3, -7) \) is **-6**.

---

### 2. **Find the slope of the equation \( y = 3x^2 - 2x \) at the point \( (1, 1) \)**

Again, we need to find the derivative of \( y = 3x^2 - 2x \), and evaluate it at \( x = 1 \).

- Derivative of \( y = 3x^2 - 2x \):
  \[
  \frac{dy}{dx} = 6x - 2
  \]
- At \( x = 1 \):
  \[
  \frac{dy}{dx} = 6(1) - 2 = 6 - 2 = 4
  \]
So, the slope at the point \( (1, 1) \) is **4**.

---

### 3. **Find the vertex of the parabola \( y = \frac{1}{8}(x^2 + 8) \)**

The equation of the parabola is given as \( y = \frac{1}{8}(x^2 + 8) \).

This is in the standard form \( y = a(x - h)^2 + k \), where \( (h, k) \) is the vertex. 

- Rewrite the equation:
  \[
  y = \frac{1}{8}x^2 + 1
  \]
Here, the term \( x^2 \) has no horizontal shift, so the vertex occurs at \( x = 0 \).
- At \( x = 0 \):
  \[
  y = \frac{1}{8}(0^2 + 8) = 1
  \]
Thus, the vertex is at \( (0, 1) \).

---

### 4. **Find the vertex of the parabola \( y = 2x^2 + 4x + 5 \)**

To find the vertex of this quadratic equation, we can use the vertex formula for a parabola \( y = ax^2 + bx + c \):
\[
x = -\frac{b}{2a}
\]
where \( a = 2 \) and \( b = 4 \).

- Calculate \( x \):
  \[
  x = -\frac{4}{2(2)} = -\frac{4}{4} = -1
  \]
- Now substitute \( x = -1 \) back into the equation to find \( y \):
  \[
  y = 2(-1)^2 + 4(-1) + 5 = 2(1) - 4 + 5 = 2 - 4 + 5 = 3
  \]
So, the vertex is at \( (-1, 3) \).

---

Would you like further details on any of the solutions or the methods used? Here are 5 follow-up questions:

1. How do we determine the slope at any given point of a curve?
2. Why do we use derivatives to find the slope of a function?
3. How can we derive the vertex formula for a quadratic equation?
4. What does the vertex of a parabola represent geometrically?
5. How can completing the square help in finding the vertex of a parabola?

**Tip**: For any quadratic equation in the form \( ax^2 + bx + c \), the vertex can be quickly found using the formula \( x = -\frac{b}{2a} \).

Find the slope of the equation y = 2 - x^2 at the point (3, -7). Find the slope of the equation y = 3x^2 - 2x at the point (1, 1). Find the vertex of the parabola y = 1/8(x^2 + 8). Find the vertex of the parabola y = 2x^2 + 4x + 5.

Learn how to find the slope of quadratic equations at specific points and determine the vertex of parabolas. This detailed step-by-step solution covers key mathematical concepts such as differentiation and quadratic functions. Ideal for high school students in Grades 10-12.

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