What is the Inverse Function of a Dropped Object's Height Over Time?

[Nelo]

Homework Statement

If an object is dropped from a height of 80m , its approximate height, h(t) metres above the ground t seconds after being dropped is given by the function h(t) = -5t^2 + 80

a) graph the functions
b) find and graph the inverse
c) is the inverse a function
d) what does the inverse represent

Homework Equations

The Attempt at a Solution

h(t) = -5t^2 + 80
t= -5h^2 +80
t-80/-5 = -5h^2
[sqrt]80 - t / 5 = h

I came up with that as the inverse function. When i graph this inverse , 80-t/5 = h do i do 80/5 , which is 16 and do [sqrt]-t+16? which is horizontal shift?

 

[Mark44]

Homework Statement

If an object is dropped from a height of 80m , its approximate height, h(t) metres above the ground t seconds after being dropped is given by the function h(t) = -5t^2 + 80

a) graph the functions
b) find and graph the inverse
c) is the inverse a function
d) what does the inverse represent

Homework Equations

The Attempt at a Solution

h(t) = -5t^2 + 80
t= -5h^2 +80
t-80/-5 = -5h^2

In the line above, you divided the left side by -5, but not the right side.

[sqrt]80 - t / 5 = h

The notation you're using above is very unhelpful. Also, you are not using parentheses, and you really need them.
The line above should be
sqrt( (80 - t)/5) = h

I came up with that as the inverse function. When i graph this inverse , 80-t/5 = h do i do 80/5 , which is 16 and do [sqrt]-t+16? which is horizontal shift?

The business of switching variable names can lead to much confusion, IMO.

If your first equation is h = -5t² + 80 = f(t),
all you really need to do is to solve for t as a function of h.

This gives you t = +sqrt( (80 - h)/5) = f^-1(h). Here I have chosen the positive square root.

The first function gives you height h as a function of t. The second function gives you time t as a function of height h. Both functions have exactly the same graph. They are two different ways of looking at the same relationship.

 

[Nelo]

ok... but, since i have whole[sqrt] of 80-t/5 = h , how do i graph that? its in the radical form like -t+80/5 but how am i supposed to go 80 units left and what od i do with the division?

Or do i simplify it by doing 80/5 and jus to the translation horizontally

 

[Mark44]

ok... but, since i have whole[sqrt] of 80-t/5 = h , how do i graph that? its in the radical form like -t+80/5 but how am i supposed to go 80 units left and what od i do with the division?

"whole[sqrt]" doesn't mean anything to me. Please use parentheses.

Your new function is h = sqrt( -1/5(t - 80))

Look at this in pieces, relative to the graph of y = sqrt(x).
How does y = sqrt(-x) change?
How does y = sqrt(-1/5 x) change?
How does y = sqrt(-1/5 (x - 80)) change?

Your original function was h = -5t² + 80. Relative to the graph of h = t², the transformations are
1) reflection across the vertical axis
2) stretch away from the horizontal axis
3) vertical shift up from the horizontal axis.

For your graph of the inverse function, which is reflected across the line y = x, what effect do these transformations have when they, too, are reflected across the line y = x?

Or do i simplify it by doing 80/5 and jus to the translation horizontally

 

[Nelo]

ok. here u go then

http://i54.tinypic.com/2ijgnww.jpg

That is the inverse eqn right?

how is that 1/5 ?

 

[Mark44]

ok. here u go then

http://i54.tinypic.com/2ijgnww.jpg

That is the inverse eqn right?

No, and you have several mistakes leading up to your final equation.

how is that 1/5 ?

It's not where it's supposed to be.

In your work, you have

t - 80 = -5h² OK
(t - 80)/(-5) = h² OK
-80/(-5) = h² Not OK - where did the t go?
(-80 + t)/(-5) = h² OK, the t reappeared.
The next line in your work has an error.

 

[Nelo]

Thats what the teacher did, so...

 

[Strants]

You need to square root the whole left side, not just the numerator. So sqrt( (80-t)/5) = h. From what you have, it looks like you take the square root of 80-t, then divide by 5, which is not the same.

 

[Mark44]

Your equation should be
h = \sqrt{\frac{80 - t}{5}}

In your work, the 5 is outside the square root. It should be inside it. If your teacher's work shows the 5 outside the radical, the teacher's work is wrong.

 

[Nelo]

In the textbook the answer says taht it is

[sqrt of all] -5t + 400/5 = t(h)

Is that wrong?

 

[Nelo]

Also, let's say that I have a function of x^2 like f(x) = 4(-2x-6) +8

Since that is able to have factors outward, and i do f(x) = 4(-2(x+6) +8

What do those numbers represent? Does the -2 still represent a horizontal compression and a reflection over the y axis?

 

[Mark44]

In the textbook the answer says taht it is

[sqrt of all] -5t + 400/5 = t(h)

Is that wrong?

I don't believe that's what your text says.

1) I'm sure it doesn't say [sqrt of all].
2) -5t + 400/5 = -5t + 80. If you mean for the 5 to divide both terms, use parentheses!
3) On the right it says t(h), which implies that the left side should have h in it, not t.

 

[Nelo]

I can scan the page if you don't believe what it says.

So , if you have something like -5 -t = 4h
and you divide -5-t/4 , the 4 doesn't multiply with the 5 and the t? creating -20-4h/4?

 

[Strants]

Do you mean
\frac{\sqrt{-5t+400}}{5}
or
\sqrt{\frac{-5t+400}{5}}

The first equation is correct, and the same as you have, though they simplified it a little, which might not be a bad idea. Try to make the denominator of the fraction a square, and then you can move it outside the root.

 

[Mark44]

Also, let's say that I have a function of x^2 like f(x) = 4(-2x-6) +8

How is this a function of x²? This is a linear polynomial function - it's graph is a straight line.

Since that is able to have factors outward, and i do f(x) = 4(-2(x+6) +8

No, you can't do this.

4(-2x - 6) + 8 = 4(-2(x + 3)) + 8 = -8(x + 3) + 8

Your expression simplifies to -8(x + 6) + 8, which is different.

Relative to y = x there are (in this order)
1) vertical stretch by a factor of 8
2) reflection across the x-axis
3) translation left by 3 units, and translation up by 8 units.

What do those numbers represent? Does the -2 still represent a horizontal compression and a reflection over the y axis?

 

[Mark44]

I can scan the page if you don't believe what it says.

So , if you have something like -5 -t = 4h
and you divide -5-t/4 , the 4 doesn't multiply with the 5 and the t? creating -20-4h/4?

PLEASE USE PARENTHESES!

-5 - t = 4h
==> (-5 - t)/4 = h

 

[Nelo]

Why is (4-x)^2 inverse not working for me.

y= (4-x)^2
x=(4-y)^2
x= -(y-4)^2
-x= -(y-4)^2
[sqrt]-x = -(y-4)
[sqrt]-x +4 = y

the answer is +- x+4 though...

 

[Mark44]

Why is (4-x)^2 inverse not working for me.

y= (4-x)^2
x=(4-y)^2
x= -(y-4)^2
-x= -(y-4)^2
[sqrt]-x = -(y-4)
[sqrt]-x +4 = y

the answer is +- x+4 though...

You're making this harder than it needs to be.
You don't need to switch variable names. Your book doesn't seem to be doing this.

There are mistakes in the 3rd, 4th, 5th, and 6th lines above.

First off, y = (4 - x)² does NOT have an inverse, since it is not a one-to-one function. If you want to get an inverse, you will have to restrict the domain so that the resulting function is one-to one. Since the graph of this function is a parabola, a reasonable restriction would be to take x only on one side of the vertex of the parabola.

Also, you can make life simpler for yourself by writing (4 - x)² as (x - 4)². The two expressions are equal, although 4 -x and x - 4 are not equal.

 

[cupid.callin]

@mark44
isn't tiny tim "2010 homework helper" award winner?

 

[Mark44]

@mark44
isn't tiny tim "2010 homework helper" award winner?

The award was given to both of us.