Question

A football is thrown into the air from an initial height of 4 feet with an upward velocity of 46 ft/sec. The function h = -16t² + 46t + 4 gives the height after t sec.

whats the vertex?

axis of symmetry?

solutions?

2 other points?

and i know the balls max height is 37.06

Answer 1

Vertex = (1.4375, 37.0625)

Axis of symmetry: t = 1.4375

x-intercept: (2.9595, 0)

y-intercept: (0, 4)

another point: (2, 32)

Explanation:

Given function:
`h(t)=-16t^2+46t+4`

(where h is the height in feet and t is the time in seconds)

The vertex is the turning point of the parabola.

To find the x-value of the turning point, differentiate the function:

`\implies h'(t)=-32t+46`

Set it to zero:

`\implies h'(t)=0`

`\implies -32t+46=0`

Solve for t:

`\implies 32t=46`

`\implies t=(23)/(16)`

Input found value of t into the function to find the y-value of the vertex:

`\implies h((23)/(16))=-16((23)/(16))^2+46((23)/(16))+4=(593)/(16)`

Therefore, the vertex is
`\left((23)/(16),(593)/(16)\right)` or (1.4375, 37.0625) in decimal form.

The axis of symmetry is the x-value of the vertex.

`\implies \textsf{Axis\:of\:Symmetry}:t=(23)/(16)=1.4375`

To find the x-intercepts, use the quadratic formula.

Quadratic Formula

`x=(-b \pm √(b^2-4ac) )/(2a)\quad\textsf{when}\:ax^2+bx+c=0`

`\implies t=(-46 \pm √(46^2-4(-16)(4)) )/(2(-16))`

`\implies t=(-46 \pm √(2372))/(-32)`

`\implies t=(23 \pm √(593))/(16)`

As time is positive,

`\implies t=(23 + √(593))/(16)=2.959474458...\:\sf s\quad only`

The y-intercept is when t = 0:

`h(0)=-16(0)^2+46(0)+4=4`

So the curve intercepts the y-axis at (0, 4)

Because of the modelling of the function, there will be a restricted domain: 0 ≤ t ≤ 2.9595

Therefore, to find another point, input a value in the domain into the function and solve:

`t=2 \implies h(2)=-16(2)^2+46(2)+4=32`

⇒ (2, 32)