Given the function $$ f(t)=2 t^{4}-6 t^{3}-8 t^{2} $$; its $$ f $$-intercept is its $$ t $$-intercepts are $$ \square $$

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To find the $$ f $$-intercept and $$ t $$-intercepts of the function $$ f(t) = 2t^4 - 6t^3 - 8t^2 $$, we can follow these steps:

### Finding the $$ f $$-intercept:
The $$ f $$-intercept occurs when $$ t = 0 $$. We can find it by evaluating $$ f(0) $$:

$$
f(0) = 2(0)^4 - 6(0)^3 - 8(0)^2 = 0
$$

Thus, the $$ f $$-intercept is $$ (0, 0) $$.

### Finding the $$ t $$-intercepts:
The $$ t $$-intercepts occur when $$ f(t) = 0 $$. We need to solve the equation:

$$
2t^4 - 6t^3 - 8t^2 = 0
$$

First, we can factor out the common term $$ 2t^2 $$:

$$
2t^2(t^2 - 3t - 4) = 0
$$

This gives us one solution:

$$
2t^2 = 0 \implies t^2 = 0 \implies t = 0
$$

Next, we need to solve the quadratic equation $$ t^2 - 3t - 4 = 0 $$. We can use the quadratic formula:

$$
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

where $$ a = 1 $$, $$ b = -3 $$, and $$ c = -4 $$:

$$
t = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-4)}}{2(1)} = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm \sqrt{25}}{2} = \frac{3 \pm 5}{2}
$$

Calculating the two possible values:

1. $$ t = \frac{3 + 5}{2} = \frac{8}{2} = 4 $$
2. $$ t = \frac{3 - 5}{2} = \frac{-2}{2} = -1 $$

Thus, the $$ t $$-intercepts are $$ t = 0, t = 4, t = -1 $$.

### Summary:
- The $$ f $$-intercept is $$ (0, 0) $$.
- The $$ t $$-intercepts are $$ t = 0, t = 4, t = -1 $$.

So, the final answer is:
- $$ f $$-intercept: $$ (0, 0) $$
- $$ t $$-intercepts: $$ 0, 4, -1 $$
The $$ f $$-intercept is $$ (0, 0) $$ and the $$ t $$-intercepts are $$ 0, 4, -1 $$.

Given the function ;
its -intercept is
its -intercepts are

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