Question

2. Complete the square to determine the vertex and axis of symmetry of y=−x2+8x+7
.

3. Find the zeros of the quadratic equation y=6x2+13x+5
. You may use any of the following methods: factoring, square roots, complete the square, or quadratic formula.

Please show all steps to solve these two questions. Upload a picture of your work for each and clearly label them as #2 and #3. You may use Desmos to check your work, but you must upload your handwritten work to receive credit

Answer 1

Sure! Below are the detailed steps to solve each problem.

### Question 2: Complete the square to determine the vertex and axis of symmetry of \( y = -x^2 + 8x + 7 \)

1. **Rewrite the equation**: Start with the quadratic equation:
\[
y = -x^2 + 8x + 7
\]

2. **Factor out the coefficient of \( x^2 \)**: Since the coefficient of \( x^2 \) is negative, factor out -1 from the first two terms:
\[
y = - (x^2 - 8x) + 7
\]

3. **Complete the square**: Take the coefficient of \( x \) (which is -8), halve it to get -4, and square it to get 16. Add and subtract this value inside the parentheses:
\[
y = - (x^2 - 8x + 16 - 16) + 7
\]
\[
y = - ((x - 4)^2 - 16) + 7
\]
\[
y = - (x - 4)^2 + 16 + 7
\]
\[
y = - (x - 4)^2 + 23
\]

4. **Identify the vertex**: The vertex form is \( y = a(x - h)^2 + k \). Here, \( h = 4 \) and \( k = 23 \), so the vertex is \( (4, 23) \).

5. **Determine the axis of symmetry**: The axis of symmetry is the vertical line that goes through the vertex. It is given by:
\[
x = h
\]
So the axis of symmetry is \( x = 4 \).

### Question 3: Find the zeros of the quadratic equation \( y = 6x^2 + 13x + 5 \)

1. **Quadratic formula**:
The zeros of the quadratic equation \( ax^2 + bx + c = 0 \) can be found using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 6 \), \( b = 13 \), and \( c = 5 \).

2. **Calculate the discriminant**:
\[
b^2 - 4ac = 13^2 - 4 \cdot 6 \cdot 5 = 169 - 120 = 49
\]

3. **Substitute into the quadratic formula**:
\[
x = \frac{-13 \pm \sqrt{49}}{2 \cdot 6}
\]
\[
x = \frac{-13 \pm 7}{12}
\]

4. **Calculate the two possible values for \( x \)**:
1. For the plus sign:
\[
x = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2}
\]
2. For the minus sign:
\[
x = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3}
\]

5. **Conclusion - zeros are**:
\[
x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3}
\]

### Summary of Results:
- **Problem 2**: The vertex is \( (4, 23) \) and the axis of symmetry is \( x = 4 \).
- **Problem 3**: The zeros of the quadratic equation are \( x = -\frac{1}{2} \) and \( x = -\frac{5}{3} \).

To fulfill your request for handwritten work, please write out the above steps clearly on paper and take a picture of your work for submission.