https://r2.erweima.ai/mathbot/image/1727074908570j67uvbod.jpg

Let's go through and solve the following equations from the image you've provided for **Problem Set 0.4**:

### 1. \( y = -x^2 + 1 \)
This is the equation of a **parabola** opening downwards, as the coefficient of \( x^2 \) is negative.

- **Symmetry**: The parabola is symmetric with respect to the y-axis.
- **x-intercepts**: Set \( y = 0 \):
  \[
  0 = -x^2 + 1 \implies x^2 = 1 \implies x = \pm 1
  \]
  So, the x-intercepts are \( (1, 0) \) and \( (-1, 0) \).
- **y-intercept**: Set \( x = 0 \):
  \[
  y = -(0)^2 + 1 = 1
  \]
  So, the y-intercept is \( (0, 1) \).

### 3. \( x = -4y^2 - 1 \)
This is a sideways parabola equation.

- **Symmetry**: This parabola is symmetric with respect to the x-axis.
- **x-intercepts**: Set \( y = 0 \):
  \[
  x = -4(0)^2 - 1 = -1
  \]
  So, the x-intercept is \( (-1, 0) \).
- **y-intercepts**: To find the y-intercept, we need to set \( x = 0 \) and solve for \( y \):
  \[
  0 = -4y^2 - 1 \implies 4y^2 = -1 + 0 \text{ (no real solutions)}
  \]
  There are **no y-intercepts**.

### 5. \( x^2 + y = 0 \)
This is the equation of a **parabola** opening downwards because \( x^2 \) is positive.

- **Symmetry**: The parabola is symmetric with respect to the y-axis.
- **x-intercepts**: Set \( y = 0 \):
  \[
  x^2 = 0 \implies x = 0
  \]
  So, the x-intercept is \( (0, 0) \).
- **y-intercept**: Set \( x = 0 \):
  \[
  y = -(0)^2 = 0
  \]
  So, the y-intercept is \( (0, 0) \), which is the same as the x-intercept.

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Would you like any more details or further clarifications on these problems? 

Here are 5 related questions to expand your understanding:

1. How do you find the vertex of a parabola from its equation?
2. Can you determine whether a parabola opens upwards or downwards from its equation?
3. What other symmetries can equations of different conic sections exhibit?
4. How do you solve for intercepts in more complex equations involving both \(x^2\) and \(y^2\)?
5. How would the equation \( y = x^2 - 1 \) differ in its graph compared to \( y = -x^2 + 1 \)?

**Tip**: For parabola equations, always look at the sign of the squared term to determine the direction in which the parabola opens.

Solve and graph quadratic equations from the problem set: y = -x^2 + 1, x = -4y^2 - 1, and x^2 + y = 0. Includes steps for finding symmetries, x- and y-intercepts, and key concepts such as parabolas and quadratic equations. Suitable for students in Grades 9-10.

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