Acannonball fired out to sea from a shore battery flows a parabolic trajectory given by the graph of the equation h(x) = 10x-0.01x^2, where h(x) is the height of the cannonball above the water when it has traveled a horizontal distance of x feet. what is the maximum height that the cannonball reaches?

since this is a parabola in the form of ax^2 + bx + c, we can find the x value of the maximum / minimum point by using the formula -b/2a, where b is the coefficient of the x term and a is the coefficient of the x^2 term.

a = -.01

b = 10

since a is negative, the graph will be head up and tails down so that the maximum / minimum point will be a maximum point.

-b/2a = -(10)/2*(-.01) = (-10)/(-.02) = 10/.02 = 500

x = 500 is the x value of the maximum point.

h(500) will be the y value.

since h(x) = 10x - .01x^2, then

h(500) = 10*500 - .01*(500)^2

simplifying:

h(500) = 5000 - .01*250000

h(500) = 5000 - 2500

h(500) = 2500

the x intercepts are when the graph of the equation crosses the x axis.

these are determined by setting the quadratic equation equal to 0.

10x - .01x^2 = 0

this can be factored to become:

x (10-.01x) = 0

x = 0

or

10-.01x = 0

solving for x in equation:

10 - .01x = 0

add .01x to both sides of equation:

10 = .01x

divide both sides of equation by .01:

10/.01 = x

simplify:

1000 = x

x intercepts are either

x = 0

or

x = 1000

this can be seen from the graph.

Step-by-step explanation:

45

step-by-step explanation:

30+15

`1234567890-=/*-qwertyuiop[]asdfghjkl; '\\zxcvbnm,./+ /