What is the Inverse of P(t)=2+10t-5t^2?

Question

And, is T(p) a function or relation? Why?

Thanks for the help!

Answer 1

P(t)= 2 + 10t - 5t^2

Keep in mind that we should have enough experience to know that this is a parabola. Parabolas will either go downward to a vertex, and then shoot upward, or go upward to a vertex, and then shoot downward.

Remember that in order to determine whether a graph is a function, it has to pass the vertical line test; that is, any vertical line will pass through AT MOST ONE point.

The "inverse" of P(t) would be a function if it passes the vertical line test, but in order for that to happen, the original function P(t) must pass the horizontal line test. We know parabolas will fail the horizontal line test (i.e. it will go through two points at some horizontal line).

Therefore, the function T(p) will be a relation.

If we wanted to solve for the inverse, we would do so as follows:

Let y = 2 + 10t - 5t^2

To solve for the inverse, swap the y variable with the t variable.

t = 2 + 10y - 5y^2

And then solve for y, by completing the square.

t = -5y^2 + 10y + 2
t = -5(y^2 - 2y) + 2
t = -5(y^2 - 2y + 1) + 2 + 5
t = -5(y^2 - 2y + 1) + 7
t = -5(y - 1)^2 + 7
t - 7 = -5(y - 1)^2

-[t - 7]/5 = (y - 1)^2

And at this point, if we solve for y, we'll obtain two values (since we have to take the square root of both sides, and then add a a "plus or minus" to the solution)

Answer 2

I would say the inverse is 1/P(t) or T(p) = 2 + 10p-5p^2 Furthermore, without doing anything, this looks like a quadratic function (more than one point answer to solve the equation- like +/-2. Functions have different answers or co-ordinates throughout the function versus relations, which have two or more inputs that can be input to derive the same output. It's best to graph the equation to view what it looks like if you're not sure. A function would be something like a straight line with a slope of +2; for every value you plug into t, you will get a different output for T(p). Whereas, a relation would be something like a parabola; for every value you plug into T(p), you could get two outputs or answers.

Answer 3

to find the inverse of P(t) exchange t with T(p) and P(t) with t
i'm going to call T(p), y for now
t = 2 +10y-5y^2
and solve for y
t-2 = 10y-5y^2
(2-t)/5 = -2y + y^2
(2-t)/5 + 1 = 1 -2y + y^2
(2-t)/5 + 1 = (-1+y)^2
+/- sqrt((2-t)/5 + 1) = -1 + y
+/- sqrt((2-t)/5 +1) + 1 = y = P^(-1)(t)
we see that T(p) has a +/- anwser so it's not a function

Answer 4

y=2+10x-5x^2

switch x and y, isolate y

x=2+10y-5y^2
7-x=5y^2-10y-2+7
7-x=5(y^2-2y+1)
(7-x)/5=(y-1)^2
y = 1 +|- sqrt ((7-x)/5)

T(p) = 1 +|- sqrt ((7-p)/5)

This is a parabola on its side, opening left, and not a function, because it fails the vertical line test.

Answer 5

UP yours man. sorry i dont know the answer.I tried to get it but it didn't work out.