Answer:
[tex]y=-5(x-0)^2+5[/tex]
Step-by-step explanation:
Given height of parabola is [tex]5\ m[/tex].
And [tex]2\ m[/tex] wide at ground level.
Also, the parabola opens down.
Let us assume the parabola is aligned on Y-axis
As the height of parabola is [tex]5\ m[/tex]. The maximum height of parabola is achieved when [tex]x=0[/tex].
So, the vertex of parabola is [tex](0,5)[/tex].
The equation of parabola having vertex [tex](h,k)[/tex] is.
[tex]y=a(x-h)^2+k[/tex].
Plugging the vertex of parabola
[tex](h,k)[/tex] [tex]=[/tex] [tex](0,5)[/tex].
[tex]h=0\ and\ k=5[/tex]
[tex]y=a(x-0)^2+5[/tex]
It is given that parabola is [tex]2\ m[/tex] wide at the ground.
As the parabola is aligned on Y-axis. So, distance between X-intercept is [tex]2\ m[/tex].
The X-intercept would be [tex](-1,0)\ and\ (1,0)[/tex]
Plugging [tex](1,0)[/tex] in the equation [tex]y=a(x-0)^2+5[/tex]
[tex]0=a(1-0)^2+5\\\\-5=a(1)^2\\\\-5=a[/tex]
Now, we get [tex]a=-5[/tex] and having vertex [tex]h=0\ and\ k=5[/tex].
So, the equation of parabola is
[tex]y=a(x-h)^2+k[/tex].
[tex]y=-5(x-0)^2+5[/tex]
