Question

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

Answer 1

To solve Question 36 and find the vertex of the quadratic equation \( y = 2x^2 + 8 \):

The vertex form of a quadratic equation is given by \( y = a(x-h)^2 + k \), where \( (h, k) \) is the vertex. For a standard form \( y = ax^2 + bx + c \), the vertex can be found using the formula:

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

In the equation \( y = 2x^2 + 8 \):

- \( a = 2 \)
- \( b = 0 \)
- \( c = 8 \)

Using the vertex formula:

\[
h = -\frac{0}{2(2)} = 0
\]

Now, substitute \( x = 0 \) back into the equation to find \( k \):

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

Therefore, the vertex is:

\[
(0, 8)
\]

### Question 36 Answer:
**Vertex: (0, 8)**

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For Question 37, we need to identify the zeros of the quadratic equation \( y = -(x - 2)^2 + 4 \).

We can find the zeros by setting \( y \) to zero and solving for \( x \):

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

Rearranging gives:

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

Taking the square root of both sides results in:

\[
x - 2 = \pm 2
\]

This gives us two equations:

1. \( x - 2 = 2 \) → \( x = 4 \)
2. \( x - 2 = -2 \) → \( x = 0 \)

Thus, the zeros are \( x = 4 \) and \( x = 0 \).

We can represent the zeros in set notation:

### Question 37 Answer:
**Zeros: {0, 4}**