Quadratic Functions Exercise 3 - Finding a Maximum of a Quadratic
The equation giving the vertical position y as a function of the
horizontal position x for a projectile launched at a 45 degree angle
with initial velocity Vo is y = (-32/Vo2)x2
+ x. If the initial velocity is Vo=20 ft/sec, what is
the maximum height that this projectile achieves? The function and
it's graph are given below. The justification has been given - no
further justification is needed.
Substitute in the given value of Vo.
The vertex of quadratic functions of the form
y=ax2 + bx + c will be at x=-b/(2a).