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

Answer:

maximum at x=5

Step-by-step 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: $x_{vertex} =-\frac{b}{2a}$

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

$x_{vertex} =-\frac{30}{2*(-3)}=\frac{30}{6} =5$

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