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

Answer:

Vertex = (1.4375, 37.0625)

Axis of symmetry:  t = 1.4375

x-intercept: (2.9595, 0)

y-intercept: (0, 4)

another point: (2, 32)

Step-by-step 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:

$\implies 32t=46$

$\implies t=\dfrac{23}{16}$

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

$\implies h(\frac{23}{16})=-16(\frac{23}{16})^2+46(\frac{23}{16})+4=\dfrac{593}{16}$

Therefore, the vertex is $\left(\dfrac{23}{16},\dfrac{593}{16}\right)$ or (1.4375, 37.0625) in decimal form.

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

$\implies \textsf{Axis\:of\:Symmetry}:t=\dfrac{23}{16}=1.4375$

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

Quadratic Formula

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when}\:ax^2+bx+c=0$

$\implies t=\dfrac{-46 \pm \sqrt{46^2-4(-16)(4)} }{2(-16)}$

$\implies t=\dfrac{-46 \pm \sqrt{2372}}{-32}$

$\implies t=\dfrac{23 \pm \sqrt{593}}{16}$

As time is positive,

$\implies t=\dfrac{23 + \sqrt{593}}{16}=2.959474458...\:\sf s\quad only$

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)