Ray needs help creating the second part of the coaster. Create a unique parabola in the pattern f(x) = (x − a)(x − b). Describe
1 answer:
You might be interested in
Answer:
Step-by-step explanation:
In order to find the horizontal distance the ball travels, we need to know first how long it took to hit the ground. We will find that time in the y-dimension, and then use that time in the x-dimension, which is the dimension in question when we talk about horizontal distance. Here's what we know in the y-dimension:
a = -32 ft/s/s
v₀ = 0 (since the ball is being thrown straight out the window, the angle is 0 degrees, which translates to no upwards velocity at all)
Δx = -15 feet (negative because the ball lands 15 feet below the point from which it drops)
t = ?? sec.
The equation we will use is the one for displacement:
Δx = and filling in:
which simplifies down to
so
so
t = .968 sec (That is not the correct number of sig fig's but if I use the correct number, the answer doesn't come out to be one of the choices given. So I deviate from the rules a bit here out of necessity.)
Now we use that time in the x-dimension. Here's what we know in that dimension specifically:
a = 0 (acceleration in this dimension is always 0)
v₀ = 80 ft/sec
t = .968 sec
Δx = ?? feet
We use the equation for displacement again, and filling in what we know in this dimension:
Δx = and of course the portion of that after the plus sign goes to 0, leaving us with simply:
Δx = (80)(.968)
Δx = 77.46 feet
that is correct hmmmm, yes, I think the restriction is also correct, it has to be x > 0, because, the domain on the original, has a restriction of x⩾3, otherwise the radicand turns negative.
now, the graph of that original function is a parabola sideways opening to the right, if you include both sides of the vertex of the parabola, it will not pass the vertical-line-test and not be a function, but if you restrict it to only positive roots, so it'll only be on the 1st quadrant, then it functions ok.
that means, the range of the original function, has to stick to only 0 or positive root values, now, recall that the range of the original is the domain of the inverse. Therefore, the domain of the inverse, will have to stick to only 0 or positive values, since it's bound to the same restriction as the original's range.
notice the picture below of both, the dashed section is the positive range for each, if you include it, the original will fail the vertical-line-test, and the inverse will fail the horizontal-line-test.
recall that for a function to have an inverse, they both must pass the vertical and horizontal line test, or be a one-to-one functions.
