Danios
contestada

A ball is shot up in the air and its height, h, above the ground in feet is given by the function h(x)= -16x^2 + 44x, where x is the number of seconds the ball has been in flight. find the maximum height that the ball attains. Round your answer to the hundredths place.

Respuesta :

GoddessofDork

So the maximum height is another word for vertex's y-coordinate. The easiest method to find the vertex is to find the axis of symmetry for the x-coordinate and then plug the axis of symmetry into the equation to solve for y.

Firstly, the equation for the axis of symmetry is [tex] x=\frac{-b}{2a} [/tex] , with b = x coefficient and a = x^2 coefficient. Using our equation, we can solve for the axis of symmetry as such:

[tex] x=\frac{-44}{2*(-16)}\\\\ x=\frac{-44}{-32} \\\\ x=1.375 [/tex]

Now that we have the x-coordinate, plug it into x and solve for y:

[tex] h(1.375)=-16*1.375^2+44*1.375\\ h(1.375)=16*1.890625+60.5\\ h(1.375)=30.25+60.5\\ h(1.375)=90.75 [/tex]

In short, the maximum height is 90.75.

MrRoyal

The maximum of a function is its vertex.

The maximum height of the ball is: 30.25 ft

The function is given as:

[tex]\mathbf{h(x) = -16x^2 + 44x}[/tex]

Differentiate

[tex]\mathbf{h'(x) = -32x + 44}[/tex]

Set to 0

[tex]\mathbf{-32x + 44 = 0}[/tex]

Collect like terms

[tex]\mathbf{-32x= -44}[/tex]

Divide both sides by -32

[tex]\mathbf{x= 1.375}[/tex]

Substitute [tex]\mathbf{x= 1.375}[/tex] in [tex]\mathbf{h(x) = -16x^2 + 44x}[/tex]

[tex]\mathbf{h(1.375) = -16(1.375)^2 + 44(1.375)}[/tex]

[tex]\mathbf{h(1.375) = 30.25}[/tex]

Hence, the maximum height of the ball is: 30.25 ft

Read more about maximum height at:

https://brainly.com/question/16602113

Otras preguntas