Question

a rocket launched from a cannon. its height, h(x), can be represented by quadratic function in terms of time, x, in seconds.

After 1 second, the rocket is 121 feet in the air; after 2 seconds, it is 224 feet in the air.

complete the height function, h(x), for this situation.

H(X)=

Answer 1

Okay I haven't done this in a while, but I think I know what to do.

Since it says it can be represented as a quadratic function, that means H(x) is in the form of ax^2 + bx + c (a, b, c are all numbers).

What I think you do is put in the coordinates in this form of an equation to create a system of equations, and then use the elimination method to find a, b, and c. So these are our three coordinate points given:
(0,0) (1, 121) and (2, 224)

Now plug these points into the quadratic formula

a(0)^2 + b(0) + c = 0
a(1)^2 + b(1) + c = 121
a(2)^2 + b(2) + c = 224

Now use elimination method to find a, b, and c now that I've given you these equations.

Answer 2

I'll give you an easy hint, c = 0.

We know this because when you multiply any number by 0 it equals 0. So by plugging in 0 for x, and since a and b are multipled by 0, that means they become 0. The equation a(0)^2 + b(0) + c = 0 is the same as c = 0.

Answer 3

I've calculated it and I know the answer, so if you need me to walk you through it just say so.

Answer 4

yeah I need you to walk me through it because I think I calculated it wrong

Answer 5

im getting something big

Answer 6

Okay, this is what I did. As I said c=0, so we can just ignore c now that we know what it is.

So we need to find a and b given these two functions
a(1)^2 + b(1) = 121 (remember I'm not writing in c anymore because we know it's 0)

and

a(2)^2 + b(2) = 224. Which is the same as 4a+ 2b = 224

We will solve for a or b by using elimination method. To use elimination method, you subtract one function from another in order to get rid of a variable.

So in my opinion I say we should get rid of b. We can do this by taking the first function and multiplying both sides by 2 so we end up getting

2a+ 2b = 242.

So now let's subtract one equation by other like this

4a + 2b = 224
-
2a + 2b = 242

We end up getting

2a = -18

Notice that b is gone, so now we only have one variable, and we can easily figure that out

a = -9

So now we know that a = -9 and c = 0. So now we can just plug in one of the original coordinate points and solve for b like this:

-9(1)^2 + b(1) = 121
-9 + b = 121
b = 130

So we end up getting a function like this as our answer

H(x) = -9x^2 + 130x

To check if it is right, you can plug in 1 for x and see if you get 121 (which you do) and plug in 2 for x and see if you get 224 (which you do). So it is correct.

Answer 7

yes that was defintely correct but you lost me on one thing that I need to understand the function part with pluging in

Answer 8

Do you mean the part right after we found that a = -9?

Answer 9

yes I got confused on that one

Answer 10

Okay sure.

So remember that quadatric equations are in the form:

y = ax^2 + bx + c. a, b, and c are just numbers. x is simply the input of the function and y is the output. Those coordinates the question gave us "After 1 second, the rocket is 121 feet in the air" just means that when you put "1" into x, y ends up equaling 121.

So what I did was point out to basically make a list out of everything in the function like this:

a = -9
b = ?
c = 0
x = 1
y = 121

Notice how the only thing we don't know here is b. So if we simply plug in 1 as x, and 121 as y, then we can use simple algebra to figure out what b is like this

-9(1)^2 + b(1) + 0 = 121
-9 + b = 121
b = 130.

If you wanted to, you could find out b by using the coordinate "after 2 seconds, it is 224 feet in the air." which is (2, 224) which is saying when x = 2, y = 224. You could plug these numbers in for x and y to solve for b as well and it would work too. I just thought using 1 for x and 121 for y was easier.

Answer 11

I apologize if I am unable to explain this to you in a way that you can understand. If my explanation is not sufficient I would recommend looking at YouTube videos on this topic. Perhaps watching a YouTube video would be more helpful to you.