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Difficulty: MediumQuadratic Functions and Graphs

An object is launched from a platform. The function h(t)=−5t2+30t+12h(t) = -5t^2 + 30t + 12 models the height of the object, in meters, tt seconds after it was launched. What is the maximum height, in meters, reached by the object?

  1. A
    3
  2. B
    12
  3. 57Answer
  4. D
    147

Answer

The maximum height reached by the object is 57 meters.
The given function h(t)=−5t2+30t+12h(t) = -5t^2 + 30t + 12 represents a parabola opening downward because the coefficient of t2t^2 is negative. The maximum value of this function occurs at its vertex. The time tt at the vertex is determined using the formula t=−b2a=−302(−5)=3t = -\frac{b}{2a} = -\frac{30}{2(-5)} = 3 seconds. Substituting t=3t = 3 back into the function yields the maximum height: h(3)=−5(3)2+30(3)+12=−45+90+12=57h(3) = -5(3)^2 + 30(3) + 12 = -45 + 90 + 12 = 57 meters.

Step-by-Step Solution

1
Identify the coefficients of the quadratic function in standard form h(t)=at2+bt+ch(t) = at^2 + bt + c.
a=−5a = -5, b=30b = 30, and c=12c = 12.
These coefficients are needed to calculate the vertex of the parabola.
2
Calculate the time tt at which the maximum height occurs using the vertex formula t=−b2at = -\frac{b}{2a}.
t=−302(−5)=3t = -\frac{30}{2(-5)} = 3 seconds.
Since the leading coefficient a=−5a = -5 is negative, the parabola opens downward, meaning its vertex represents the maximum value.
3
Substitute t=3t = 3 back into the height function to find the maximum height.
h(3)=−5(3)2+30(3)+12=−45+90+12=57h(3) = -5(3)^2 + 30(3) + 12 = -45 + 90 + 12 = 57 meters.
Evaluating the function at the time of the vertex gives the corresponding maximum height.

Key Concept

The maximum value of a quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + c (where a<0a < 0) is the vertical coordinate of its vertex, which occurs at x=−b2ax = -\frac{b}{2a}.
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Topics

Quadratic Functions and Graphs

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