Algebra word problem

[amathproblemthatneedsolve]

A sea diving bird dives off a cliff into the sea to catch a herring. Its vertical height h above the sea level in metres is given by the equation h = t 3 – 6t 2 – t + 30, 0 ≤ t < 6 where t is the time in seconds since the bird dived off the cliff.

a. Show that this cubic equation can be written as h = (t + 2)(t – 3)(t – 5)

b. At what times is the bird at sea level?

c. How long is the bird underwater?

At the moment I'm stumped and need some guidance, as there are no other examples like this one to go by. Thanks

 

[MarkFL]

I am assuming we are given:

[MATH]h(t)=t^3-6t^2-t+30[/MATH]
For part (a)...can you show that [MATH]h(-2)=h(3)=h(5)=0[/MATH]? If so, what does this imply?

 

[amathproblemthatneedsolve]

h=t³-6t²-t+30 Don't know if thats the same as you put

 

[MarkFL]

h=t³-6t²-t+30 Don't know if thats the same as you put

It looks the same to me. Can you answer my question?

 

[amathproblemthatneedsolve]

I'm not totally sure how to show h(−2)=h(3)=h(5)=0

 

[MarkFL]

Let me put it another way...when \(t=-2\), what is \(h\)?

 

[amathproblemthatneedsolve]

For (a) can't I just show that if I work h = (t + 2)(t – 3)(t – 5) out I get h = t ³ – 6t ² – t + 30 so therfore h = t ³ – 6t ² – t + 30 can be written as h = (t + 2)(t – 3)(t – 5)

 

[MarkFL]

For (a) can't I just show that if I work h = (t + 2)(t – 3)(t – 5) out I get h = t ³ – 6t ² – t + 30 so therfore h = t ³ – 6t ² – t + 30 can be written as h = (t + 2)(t – 3)(t – 5)

Yes, but the path to which I am pointing is quicker. Given a polynomial function \(f(x)\), if we can show that \(f(a)=0\), then we know \(x-a\) must be a factor of \(f(x)\).

 

[Otis]

h=t³-6t²-t+30 Don't know if thats the same as you put

Hello. MarkFL used function notation. Your comment (in red above) led me to wonder whether you're familiar with function notation, yet.

When you see expressions like h(t) and h(2), do you think they mean h×t and h×2? (They don't; it's function notation.)

If your class has been talking about the relationship between a polynomial's roots and its factors, then you may be expected to use that method, for part (a). Mark posted the relationship, in post #8.

Have you learned the basic shape of a cubic polynomial curve? If so, then you know that only a portion of the polynomial's graph is used, in this exercise. For example, there are no negative values of t; time starts when the bird dives at t=0. (I note also that they did not extend the graph all the way to t=6, but I guess that's okay because they haven't asked any questions about that part.)

Can you determine the value of t, at the bird's entry point? How about at the resurface point? Those times allow you to answer both parts (b) and (c).

?

 

[Otis]

... h = t 3 – 6t 2 – t + 30 ...

A quick note about typing exponents. We use the caret symbol ^ for that.

h = t^3 – 6t^2 – t + 30

Cheers
\(\;\)

 

[amathproblemthatneedsolve]

I'll read up about this and get back to both of you!!

 

[amathproblemthatneedsolve]

Okay so I found out that they only want me to expand h = (t + 2)(t – 3)(t – 5) into h = t^3 – 6t^2 – t + 30 and that suffices for a.

for b. I'm thinking we use 0<t<6 how do we use this?

 

[Otis]

... for [part b] I'm thinking we use 0<t<6

In post #1, you wrote 0 ≤ t < 6 (in other words, 0 is included).

That compound inequality gives us the time interval in this exercise. The value of t tells us how many seconds have elapsed since the bird dove from the clifftop. For part (b), that interval tells us only that negative values of t need to be ignored.

Part (b) asks for the values of t when the bird is at sea level. The bird is at sea level when its height is h=0 meters. That is, you need to find the positive values of t where the graph crosses the horizontal axis.

You've been taught about x-intercepts before, and how to calculate them from a factored polynomial, yes? Part (b) simply asks for the positive intercepts, except in this exercise they're using symbol t instead of x (so the horizontal axis is called the t-axis and the points on it where h=0 are called t-intercepts instead of x-intercepts).

Find the values of t that make t^3–6t^2–t+30 equal zero. In other words, what is the value of t at the entry point? What is the value of t at the resurface point?

If you forgot how to find x-intercepts, then here are the steps for part (b):

1) Set each of the polynomial's factors equal to zero
2) Solve each of those three equations for t
3) Ignore any negative results

?

 

[amathproblemthatneedsolve]

@Otis My apologies for the late reply as I had an issue with logging in. So 0= t^3–6t^2–t+30 with the t= results there are only two because one is a negative answer (t=-2) so t=3 and t=5

 

[Otis]

... so t=3 and t=5

That's the correct answer for part (b). You could also answer with a sentence (including units):

"The bird was at sea level at t = 3 sec and t = 5 sec."

Looking at the graph, we can see that the bird was underwater from t=3 through t=5. Now you know the answer to part (c).

?

 

[amathproblemthatneedsolve]

@Otis so all parts are now solved?

 

[Otis]

... all parts are now solved?

It seems so. You answered part (a) in post #12. You answered part (b) in post #14.

I'm thinking you know how many seconds the bird was underwater, so you can answer part (c), and all posted parts will have been solved by you.

If you have any remaining questions about this exercise, feel free to post them.

?

 

[amathproblemthatneedsolve]

@Otis is the bird underwater for 5-3 2sec? Because The birdt enters the water when t = 3 and leaves the water when t = 5. The bird is under water for 5 − 3 = 2 sec.

 

[amathproblemthatneedsolve]

or @MarkFL

 

[JeffM]

@Otis is the bird underwater for 5-3 2sec? Because The birdt enters the water when t = 3 and leaves the water when t = 5. The bird is under water for 5 − 3 = 2 sec.

Yes.