Ithe vertex, $$ x $$ and $$ y $$ intercepts of the function (write in space below) and graph the function on graph provided. You may leave the answer blank empty. 5) $$ f(x)=x^{2}-4 x-12 $$ Vertex: $$ (1,-19 $$ $$ y $$ intercept: $$ (0,12) $$

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To find the vertex and intercepts of the function $$ f(x) = x^2 - 4x - 12 $$, we can follow these steps:

### 1. Vertex
The vertex of a quadratic function in the form $$ f(x) = ax^2 + bx + c $$ can be found using the formula:
$$
x = -\frac{b}{2a}
$$
For the function $$ f(x) = x^2 - 4x - 12 $$:
- $$ a = 1 $$
- $$ b = -4 $$
- $$ c = -12 $$

Calculating the x-coordinate of the vertex:
$$
x = -\frac{-4}{2 \cdot 1} = \frac{4}{2} = 2
$$

Now, substitute $$ x = 2 $$ back into the function to find the y-coordinate:
$$
f(2) = (2)^2 - 4(2) - 12 = 4 - 8 - 12 = -16
$$
So, the vertex is $$ (2, -16) $$.

### 2. Y-Intercept
The y-intercept occurs when $$ x = 0 $$:
$$
f(0) = (0)^2 - 4(0) - 12 = -12
$$
Thus, the y-intercept is $$ (0, -12) $$.

### 3. X-Intercepts
To find the x-intercepts, we set $$ f(x) = 0 $$:
$$
x^2 - 4x - 12 = 0
$$
We can factor this quadratic:
$$
(x - 6)(x + 2) = 0
$$
Setting each factor to zero gives:
$$
x - 6 = 0 \quad \Rightarrow \quad x = 6
$$
$$
x + 2 = 0 \quad \Rightarrow \quad x = -2
$$
Thus, the x-intercepts are $$ (6, 0) $$ and $$ (-2, 0) $$.

### Summary
- Vertex: $$ (2, -16) $$
- Y-Intercept: $$ (0, -12) $$
- X-Intercepts: $$ (6, 0) $$ and $$ (-2, 0) $$

You can now graph the function using these points. The parabola opens upwards since the coefficient of $$ x^2 $$ is positive.
Vertex: $$ (2, -16) $$ $$ y $$ intercept: $$ (0, -12) $$ X-intercepts: $$ (6, 0) $$ and $$ (-2, 0) $$

Ithe vertex, and intercepts of the function (write in space below) and graph the function on
graph provided. You may leave the answer blank empty.
5)
Vertex:
intercept:

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Ithe vertex, and intercepts of the function (write in space below) and graph the function on graph provided. You may leave the answer blank empty. 5) Vertex: intercept: