A marine biologist is testing a new under water probe. The probe is catapulted from the ship into the air, descends under water to a predetermined depth, and then returns to the surface where it can be collected. The function that models the probe’s path is f(x) = 15t2 – 195t + 450, where t is time measured in seconds. The probe enters the water at 3 seconds. How long is the probe under water?

Well, it seems like the probe is having quite the adventure! Let's find out how long it spends underwater.

To do that, we need to determine when the probe is at the water's surface, which corresponds to the zeros of the function. In other words, we need to find the values of t for which f(t) equals zero.

So, let's set the function to zero and solve for t:

0 = 15t^2 - 195t + 450

Now, we could solve this equation using various methods like factoring, completing the square, or the quadratic formula. But I'm feeling a bit like a daredevil today, so let's use the quadratic formula!

The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by:

x = (-b ± √(b^2 - 4ac)) / 2a

In our case, a = 15, b = -195, and c = 450. Plugging these values into the formula will give us the values of t when the probe is at the water's surface.

Before I do the math though, remember to always check if there's any sharks or submarines nearby when retrieving the probe!

Calculating... calculating... it seems that the probe is underwater for a total of 6 seconds! That's quite a dip for our little adventurer.

Just make sure to dry it off properly before you collect it, wouldn't want any soggy data.

To determine how long the probe is underwater, we need to find the range of values of t for which the probe is at a depth below the surface of the water.

The depth of the probe is given by the function f(x) = 15t^2 - 195t + 450.

We know that the probe enters the water at 3 seconds. Therefore, we need to find the values of t for which the depth f(t) is non-negative.

To do this, we set f(t) = 0 and solve for t:

0 = 15t^2 - 195t + 450

Using factoring or the quadratic formula, we can solve this equation. Let's use the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac)) / (2a)

In this equation, a = 15, b = -195, and c = 450.

Plugging in these values, we get:

t = (-(-195) ± sqrt((-195)^2 - 4(15)(450))) / (2*15)
= (195 ± sqrt(38025 - 27000)) / 30
= (195 ± sqrt(11025)) / 30
= (195 ± 105) / 30

Now, we have two solutions for t:

t₁ = (195 + 105) / 30 = 10
t₂ = (195 - 105) / 30 = 3/2

Since the probe enters the water at 3 seconds, we discard t₂ as a solution. Therefore, the probe is underwater from t = 3 seconds to t = 10 seconds.

Hence, the probe is underwater for 10 - 3 = 7 seconds.

To find out how long the probe is under water, we need to determine the values of t when the probe is below the water surface.

The probe enters the water at 3 seconds. To determine how long the probe is under water, we need to find the values of t for which the function f(t) equals or exceeds the depth of the water's surface.

The depth of the water's surface can be represented by f(t) = 0, since the probe enters and exits at the surface.

To find the values of t, we can set f(t) = 0 and solve for t:

0 = 15t^2 - 195t + 450

To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, factoring may not be straightforward, so we'll use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 15, b = -195, and c = 450.

Using the quadratic formula, we can solve for t:

t = (-(-195) ± √((-195)^2 - 4 * 15 * 450)) / (2 * 15)
= (195 ± √(38025 - 27000)) / 30
= (195 ± √(11025)) / 30
= (195 ± 105) / 30

Now we have two possible values for t:

t1 = (195 + 105) / 30 = 300 / 30 = 10
t2 = (195 - 105) / 30 = 90 / 30 = 3

From these values, we can see that the probe is under water from t = 3 to t = 10 seconds.

Therefore, the probe is under water for 10 - 3 = 7 seconds.

f(t) = 15(t-3)(t-10)

Clearly f(t) < 0 on the interval (3,10)