A projectile follows a parabolic path whose height, in meters, is given by the function f(x) = -x^2 + 2x + 2. Find the maximum horizontal distance that the p

Question

Please answer the above question with steps and explanation

 

Answer 1

Assume the parabola is on the x-axis and y-axis.

The x-axis is the ground.

The parabola opens down.

Find the two points where the parabola cuts the x-axis.

When the parabola cuts the x-axis, the value of y is 0; in other words, f(x)=0

first of all, what is the vertex of the parabola ?

use -b/2a

-2/-2=1 for x

y=-1²+2(1)+2

y=-1+2+2=3 for y

the vertex is (1,3)

now find where the parabola intercepts the x-axis

0=-x²+2x+2

using the quadratic equation...

[-2+√(4-4(-1)(2)]/2(-1)

[-2+√12]/-2

[-2+2√3]/-2

1+√3

1+1.73205 and 1-1.73205

2.73205 and -0.73205 are the points where the parabola intercepts the x-axis

(2.73205,0) and (-0.73205,0) to be more specific

on the x-axis...

what is the distance from -0.73205 to 2.73205 on the x-axis ?

add the absolute values of the two numbers

l-0.73205l+l2.73205l=0.73205+2.73205=3.4641 meters is the maximum horizontal distance