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: [tex]h(t)=-16t^2+46t+4[/tex]
(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:
[tex]\implies h'(t)=-32t+46[/tex]
Set it to zero:
[tex]\implies h'(t)=0[/tex]
[tex]\implies -32t+46=0[/tex]
Solve for t:
[tex]\implies 32t=46[/tex]
[tex]\implies t=\dfrac{23}{16}[/tex]
Input found value of t into the function to find the y-value of the vertex:
[tex]\implies h(\frac{23}{16})=-16(\frac{23}{16})^2+46(\frac{23}{16})+4=\dfrac{593}{16}[/tex]
Therefore, the vertex is [tex]\left(\dfrac{23}{16},\dfrac{593}{16}\right)[/tex] or (1.4375, 37.0625) in decimal form.
The axis of symmetry is the x-value of the vertex.
[tex]\implies \textsf{Axis\:of\:Symmetry}:t=\dfrac{23}{16}=1.4375[/tex]
To find the x-intercepts, use the quadratic formula.
Quadratic Formula
[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when}\:ax^2+bx+c=0[/tex]
[tex]\implies t=\dfrac{-46 \pm \sqrt{46^2-4(-16)(4)} }{2(-16)}[/tex]
[tex]\implies t=\dfrac{-46 \pm \sqrt{2372}}{-32}[/tex]
[tex]\implies t=\dfrac{23 \pm \sqrt{593}}{16}[/tex]
As time is positive,
[tex]\implies t=\dfrac{23 + \sqrt{593}}{16}=2.959474458...\:\sf s\quad only[/tex]
The y-intercept is when t = 0:
[tex]h(0)=-16(0)^2+46(0)+4=4[/tex]
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:
[tex]t=2 \implies h(2)=-16(2)^2+46(2)+4=32[/tex]
⇒ (2, 32)
