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An athlete to event is the shot put releases a shot. When the shoot whose path is shown by the graph to the right is released of an angle of 40° its height f(x) in feet can be modeled by f(x) =-0.01x^2+0.8x +5.7 where X is the SHOTS horizontal distance in feet from its point of release use the model to Solve parts a through C and verify your answer through the graph

An athlete to event is the shot put releases a shot. When the shoot whose path is-example-1
User Fejta
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1 Answer

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Explanation

Given information


f(x)=-0.01x^2\text{ + 0.8x + 5.7}

Step1: Find the value of x from the above quadratic function

To find the value of x we need to make f(x) = 0


0=-0.01x^2\text{ + 0.8x + 5.7}

Step 2: Apply the general quadratic rule to find the value of x


x\text{ = }\frac{-b\text{ }\pm\sqrt[]{b^2\text{ - 4ac}}}{2a}

From the given function;

a = -0.01

b = 0.8

c = 5.7


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User Ziv Galili
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