Question

NO LINKS!! A quarterback tosses a football to a receiver 40 yards downfield. The height of the football, f(x), in feet, can be modeled by f(x)= -0.0.25x+x+6, where x is the ball's horizontal distance, in yards, from the quarterback. What is the ball's maximum height and how far from the quarterback does this occur?​

Answer 1

Answer:

• maximum height: 16 yards
• distance from quarterback: 20 yards

Step-by-step explanation:

We assume the height model is supposed to be ...

  h(x) = -0.025x² +x +6

The maximum height is found at the vertex of the parabolic curve described by this equation. For standard form expression ax² +bx +c, the vertex is found at ...

  x = -b/(2a)

  x = -1/(2(-0.025)) = 1/0.05 = 20

The ball will reach its maximum height 20 yards from the quarterback.

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That height can be found by evaluating the function for x=20:

  h(20) = -0.025(20²) +20 +6 = -10 +20 +6 = 16

The ball's maximum height is 16 yards.

Answer 2

Answer:

Maximum height = 16 yards

This occurs 20 yards from the quarterback.

Step-by-step explanation:

Given function:  $f(x)=-0.025x^2+x+6$

To find the maximum (turning point), differentiate the function:

$f('x)=-0.05x+1$

Set the derivative to zero and solve for $x$:

$\implies f'(x)=0$

$\implies -0.05x+1=0$

$\implies 0.05x=1$

$\implies x=20$

Substitute found value of $x$ into the function to find the maximum height:

$f(20)=-0.025(20)^2+(20)+6=16$

Therefore, the maximum height is 16 yards.

This occurs 20 yards from the quarterback.