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.

(6 points)

Answer 1

Let's solve both problems step by step.

### Problem 2: Complete the square for \( y = -x^2 + 8x + 7 \)

**Step 1: Factor out the negative sign from the quadratic terms**

\[
y = - (x^2 - 8x) + 7
\]

**Step 2: Complete the square inside the parentheses**

To complete the square, we need to take half of the coefficient of \( x \) (which is -8), square it and add and subtract that value inside the parentheses.

Half of -8 is -4. Squaring -4 gives us 16.

So, we rewrite the expression as follows:

\[
y = - (x^2 - 8x + 16 - 16) + 7
\]

This simplifies to:

\[
y = - ( (x - 4)^2 - 16) + 7
\]

**Step 3: Distribute the negative sign**

Now, distribute the negative sign and simplify:

\[
y = - (x - 4)^2 + 16 + 7
\]
\[
y = - (x - 4)^2 + 23
\]

**Step 4: Identify the vertex and axis of symmetry**

The vertex form of a quadratic function is \( y = a(x-h)^2 + k \), where \( (h, k) \) is the vertex.

From our equation, we have:

- Vertex \( (h, k) = (4, 23) \)
- Axis of symmetry: \( x = h = 4 \)

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

**Step 1: Apply the quadratic formula**

The quadratic formula is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For our quadratic \( ax^2 + bx + c \) where \( a = 6 \), \( b = 13 \), and \( c = 5 \):

**Step 2: Calculate the discriminant \( b^2 - 4ac \)**

\[
b^2 = 13^2 = 169
\]
\[
4ac = 4(6)(5) = 120
\]
\[
b^2 - 4ac = 169 - 120 = 49
\]

**Step 3: Substitute values into the quadratic formula**

Since the discriminant is positive, we have two distinct real roots.

\[
x = \frac{-13 \pm \sqrt{49}}{2 \cdot 6} = \frac{-13 \pm 7}{12}
\]

**Step 4: Calculate the two solutions**

1. Using \( + \):
\[
x = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2}
\]

2. Using \( - \):
\[
x = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3}
\]

**Final Roots:**
The zeros are \( x = -\frac{1}{2} \) and \( x = -\frac{5}{3} \).

---

Now for handwriting, you will need to write these steps clearly on paper. You can upload the handwritten work as images labeled as #2 for the vertex and axis of symmetry and #3 for the zeros of the quadratic equation.

If you need further assistance or a summary, feel free to ask!