Question

Question 35
Find the axis of symmetry of the following quadratic equation. Answer with x=
*y=-x^-4x+1
1 point
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Question 36 Find the vertex of the following quadratic equation. Answer as an ordered pair (x,y)
*y=2x^2+8
1 point
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Question 37
Identify the zeros of the following quadratic equation. Answer in series notation { }
*y=-(x-2)^2+4
1 point
Captionless Image

Answer 1

Let's solve each question one by one.

### Question 35: Find the axis of symmetry for the equation \( y = -x^2 - 4x + 1 \)

The general form of a quadratic equation is \( y = ax^2 + bx + c \). The axis of symmetry can be found using the formula:

\[
x = -\frac{b}{2a}
\]

In the equation \( y = -x^2 - 4x + 1 \):
- \( a = -1 \)
- \( b = -4 \)

Plugging these values into the formula:

\[
x = -\frac{-4}{2 \cdot -1} = \frac{4}{-2} = -2
\]

So the axis of symmetry is:
\[
x = -2
\]

### Question 36: Find the vertex of the equation \( y = 2x^2 + 8 \)

For a quadratic in the form \( y = ax^2 + bx + c \), the vertex can also be found using the axis of symmetry. The x-coordinate of the vertex is given by:

\[
x = -\frac{b}{2a}
\]

In this equation:
- \( a = 2 \)
- \( b = 0 \)

Calculating the x-coordinate of the vertex:

\[
x = -\frac{0}{2 \cdot 2} = 0
\]

To find the y-coordinate, substitute \( x = 0 \) into the equation:

\[
y = 2(0)^2 + 8 = 8
\]

Thus, the vertex is:
\[
(0, 8)
\]

### Question 37: Identify the zeros of the equation \( y = -(x-2)^2 + 4 \)

To find the zeros, we set \( y \) to zero and solve for \( x \):

\[
0 = -(x-2)^2 + 4
\]
\[
-(x-2)^2 = -4
\]
\[
(x-2)^2 = 4
\]

Taking the square root of both sides gives:

\[
x - 2 = 2 \quad \text{or} \quad x - 2 = -2
\]

Solving each equation:

1. \( x - 2 = 2 \) leads to \( x = 4 \)
2. \( x - 2 = -2 \) leads to \( x = 0 \)

Thus, the zeros are:
\[
\{0, 4\}
\]

### Summary of Answers:

- **Question 35**: Axis of symmetry: \( x = -2 \)
- **Question 36**: Vertex: \( (0, 8) \)
- **Question 37**: Zeros: \( \{0, 4\} \)