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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 V_{o} is y = (-32/V_{o}^{2})x^{2}
+ x. If the initial velocity is V_{o}=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.

STATEMENT

JUSTIFICATION

Given

Substitute in the given value of V_{o}.

The vertex of quadratic functions of the form
y=ax^{2} + bx + c will be at x=-b/(2a).

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

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