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Quadratic Functions Exercise 3 - Finding a Maximum of a Quadratic Function

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).


The maximum y-value occurs at the vertex x-value.











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