Someone help me please.

Step-by-step explanation:

f(x) = -0.01x² + 0.7x + 6.1

a) f(x) is a downward facing parabola, so the maximum height is at the vertex.  The vertex of a parabola can be found using x = -b/(2a).

x = -0.7 / (2 × -0.01)

x = 35

f(35) = -0.01(35)² + 0.7(35) + 6.1

f(35) = 18.35

The maximum height is 18.35 feet.

b) The maximum horizontal distance is when the ball lands, or when f(x) = 0.

0 = -0.01x² + 0.7x + 6.1

0 = x² − 70x − 610

Solve with quadratic formula:

x = [ -b ± √(b² − 4ac) ] / 2a

x = [ -(-70) ± √((-70)² − 4(1)(-610)) ] / 2(1)

x = (70 ± √7340) / 2

x = 35 ± √1835

x = -7.84, 77.84

x can't be negative, so x = 77.84.  The ball's maximum horizontal distance is 77.84 feet.

c) When the ball is first launched, x = 0.  The height at that position is:

f(0) = 6.1

The ball is launched from an initial height of 6.1 feet.

true? is that what you're asking or do you mean something else?