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an athlete whose event is the shot put releases a shot. when the shot w…

Question

an athlete whose event is the shot put releases a shot. when the shot whose path is shown by the graph to the right is released at an angle of 35°, its height, f(x), in feet, can be modeled by f(x)= -0.01x² + 0.7x + 5.7, where x is the shots horizontal distance, in feet, from its point of release. use this model to solve parts (a) through (c) and verify your answers using the graph.
a. what is the maximum height of the shot and how far from its point of release does this occur?
the maximum height is □, which occurs □ feet from the point of release.
(type an integer or decimal rounded to four decimal places as needed.)

Explanation:

The function \( f(x) = -0.01x^2 + 0.7x + 5.7 \) is a quadratic function in the form \( f(x)=ax^2 + bx + c \), where \( a=-0.01 \), \( b = 0.7 \), and \( c = 5.7 \). For a quadratic function, the vertex represents the maximum (if \( a<0 \)) or minimum (if \( a>0 \)) point. The x - coordinate of the vertex of a quadratic function \( f(x)=ax^2+bx + c \) is given by \( x=-\frac{b}{2a} \), and we can find the y - coordinate (the maximum height in this case) by substituting this x - value back into the function.

Step 1: Find the x - coordinate of the vertex (horizontal distance for maximum height)

We know that for \( f(x)=-0.01x^2 + 0.7x + 5.7 \), \( a=-0.01 \) and \( b = 0.7 \).
Using the formula for the x - coordinate of the vertex \( x=-\frac{b}{2a} \)
Substitute \( a=-0.01 \) and \( b = 0.7 \) into the formula:
\( x=-\frac{0.7}{2\times(-0.01)} \)
First, calculate the denominator: \( 2\times(-0.01)=-0.02 \)
Then, \( x =-\frac{0.7}{-0.02}=\frac{0.7}{0.02}=35 \)

Step 2: Find the y - coordinate of the vertex (maximum height)

Now that we know the x - coordinate of the vertex is \( x = 35 \), we substitute \( x = 35 \) into the function \( f(x)=-0.01x^2+0.7x + 5.7 \)
\( f(35)=-0.01\times(35)^2+0.7\times35 + 5.7 \)
First, calculate \( (35)^2 = 1225 \), then \( -0.01\times1225=-12.25 \)
Next, calculate \( 0.7\times35 = 24.5 \)
Now, substitute these values back into the function:
\( f(35)=-12.25 + 24.5+5.7 \)
First, \( -12.25+24.5 = 12.25 \)
Then, \( 12.25 + 5.7=17.95 \)

Answer:

The maximum height is \( 17.95 \), which occurs \( 35 \) feet from the point of release.