The flight is in the shape of a parabola with a vertex 5 feet above the water and 1/2 * 33 = 16.5 feet horizontally from the point of leaving the water

y = a(x - h)^2 + k

where (h,k) is the vertex of the parabola and here it is (5 , 16.5), so we have the function:-

y = a(x - 16.5)^2 + 5

when x = 0 y = 0 so

0 = a(-16.5)^2 + 5

which gives a = -0.018365

So our function for the flight path is

y = -0.018365(x - 16.5)^2 + 5 Answer

7 0

3 years ago

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Evaluate the difference quotient for the given function. Simplify your answer.

nasty-shy [4]

I suppose you mean

$f(x) = \dfrac{x+5}{x+1}$

Then

$f(3) = \dfrac{3+5}{3+1} = \dfrac84 = 2$

and the difference quotient is

$\dfrac{f(x)-f(3)}{x-3} = \dfrac{\frac{x+5}{x+1}-2}{x-3} \\\\ \dfrac{f(x)-f(3)}{x-3} = \dfrac{\frac{x+5-2(x+1)}{x+1}}{x-3} \\\\ \dfrac{f(x)-f(3)}{x-3} = \dfrac{-x+3}{(x+1)(x-3)} \\\\ \dfrac{f(x)-f(3)}{x-3} = \boxed{-\dfrac{x-3}{(x+1)(x-3)}}$

If it's the case that x ≠ 3, then (x - 3)/(x - 3) reduces to 1, and you would be left with

$\dfrac{f(x)-f(3)}{x-3}\bigg|_{x\neq3} = -\dfrac1{x+1}$

4 0

2 years ago