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: [tex]y(x)=ax^2+bx+c[/tex],
the x-value of its vertex is given by the expression: [tex]x_{vertex} =-\frac{b}{2a}[/tex]
In our case, (since [tex]a[/tex]= -3, and [tex]b[/tex]= 30) it becomes:
[tex]x_{vertex} =-\frac{30}{2*(-3)}=\frac{30}{6} =5[/tex]
Therefore, the projectile will reach its maximum at x=5
