Question

A projectile is shot into the air following the path, h(x) = -3x^2 + 30x + 300. At what time, value of x, will it reach a maximum height?

x = 1
x = 2
x = 4
x = 5

Answer 1

maximum at x=5

Explanation:

This path is parabolic (represented by a parabola since it is given by a quadratic expression).The parabola has the branches down since its leading coefficient is negative (-3), so the maximum value of this function will correspond to the vertex of the parabola.

We can use the definition for the vertex of a parabola to find it.

For a general parabola of the form:
`y(x)=ax^2+bx+c`,

the x-value of its vertex is given by the expression:
`x_(vertex) =-(b)/(2a)`

In our case, (since
`a`= -3, and
`b`= 30) it becomes:

`x_(vertex) =-(30)/(2*(-3))=(30)/(6) =5`

Therefore, the projectile will reach its maximum at x=5