Question

For each of the following quadratic functions (Show your work):
Calculate the vertex by hand and write it as an ordered pair.
Determine the axis of symmetry and write it as a linear equation (x = # or t = #).
Function p(t)=5t^2+20t
-b/ 2a
f(-b/2a)
Vertex
Axis of Symmetry (Be sure to use the proper variable)

Answer 1

To find the vertex of the quadratic function \( p(t) = 5t^2 + 20t \), we will use the formula for the vertex of a quadratic function, which is given by:

### Step 1: Find the vertex using \( t = -\frac{b}{2a} \)

Here, the coefficients are:
- \( a = 5 \)
- \( b = 20 \)

Using the vertex formula:

\[
t = -\frac{b}{2a} = -\frac{20}{2(5)} = -\frac{20}{10} = -2
\]

### Step 2: Calculate \( p(-2) \)

Next, we need to substitute \( t = -2 \) back into the function to find the \( p(t) \) value at this \( t \):

\[
p(-2) = 5(-2)^2 + 20(-2)
\]
\[
= 5(4) + 20(-2)
\]
\[
= 20 - 40
\]
\[
= -20
\]

### Step 3: Write the vertex

Now, we can write the vertex as an ordered pair:

\[
\text{Vertex} = (-2, -20)
\]

### Step 4: Determine the axis of symmetry

The axis of symmetry is the vertical line that passes through the vertex. Since we found that the vertex is at \( t = -2 \), the equation of the axis of symmetry is:

\[
\text{Axis of Symmetry: } t = -2
\]

### Summary of Results

- **Vertex:** \((-2, -20)\)
- **Axis of Symmetry:** \( t = -2 \)