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 h(t)=5t^2
-b/2a
f(-b/2a)
vertex
axis of symmetry

Answer 1

To find the vertex and axis of symmetry of the quadratic function \( h(t) = 5t^2 \), we need to follow these steps:

### Step 1: Identify coefficients
In the standard form of a quadratic function, \( h(t) = at^2 + bt + c \):

- \( a = 5 \)
- \( b = 0 \)
- \( c = 0 \)

### Step 2: Calculate the vertex using \(-\frac{b}{2a}\)
The x-coordinate (or in this case, t-coordinate) of the vertex can be found using the formula:

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

Substituting the values of \( a \) and \( b \):

\[
t = -\frac{0}{2 \cdot 5} = 0
\]

### Step 3: Calculate the function value at the vertex
Next, we calculate the value of the function \( h(t) \) at \( t = 0 \):

\[
h(0) = 5(0)^2 = 0
\]

### Step 4: Write the vertex as an ordered pair
Thus, the vertex of the function is:

\[
(0, 0)
\]

### Step 5: Determine the axis of symmetry
The axis of symmetry can be written as a linear equation, which corresponds to the x-coordinate of the vertex:

\[
t = 0
\]

### Summary of Results
- **Vertex**: \( (0, 0) \)
- **Axis of Symmetry**: \( t = 0 \)

These steps show the calculations of the vertex and axis of symmetry for the quadratic function \( h(t) = 5t^2 \).