Question

The equation P(x)=−x2+50x+500 models the height of a projectile, in feet, after x seconds. After how many seconds will the projectile reach its maximum height?

Answer 1

25 seconds

Explanation:

we have

`P(x)=-x^(2)+50x+500`

This is the equation of a vertical parabola open downward

The vertex is a maximum

The y-coordinate of the vertex is the maximum height of the projectile

The x-coordinate of the vertex is the time in seconds that the projectile takes to reach its maximum height.

Convert the quadratic equation in vertex form

we have

`P(x)=-x^(2)+50x+500`

Factor -1

`P(x)=-(x^(2)-50x)+500`

Complete the square

`P(x)=-(x^(2)-50x+25^2)+500+25^2`

`P(x)=-(x^(2)-50x+625)+500+625`

`P(x)=-(x^(2)-50x+625)+1,125`

Rewrite as perfect squares

`P(x)=-(x-25)^(2)+1,125`

The vertex is the point (25,1,125)

therefore

The maximum height of the projectile is 1,125 feet

The time in seconds that the projectile takes to reach its maximum height is 25 seconds