Answer:
1. The correct option B.
2.The correct option A.
Step-by-step explanation:
The given function is
[tex]f(d)=-0.6d^2+5.4d+0.8[/tex]
Where f(d) is the height of the ball at horizontal distance d.
Put f(d)=0, to find the distance where the ball touch the ground.
[tex]0=-0.6d^2+5.4d+0.8[/tex]
Quadratic formula:
[tex]d=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Using the quadratic formula we get
[tex]d=\frac{-5.4\pm \sqrt{(5.4)^2-4(-0.6)(0.8)}}{2(-0.6)}[/tex]
[tex]d=-0.146,9.146[/tex]
Therefore the ball is in air between d=-0.146 to d=9.146.
The distance can not be negative, therefore the ball remains in the air between d=0 to d=9.146.
[tex]9.146\approx 9.15[/tex]
Therefore the correct option is B.
2.
The given equation is
[tex]y=(x-1)^2+1[/tex] .... (1)
The standard form of parabola is
[tex]y=a(x-h)^2+h[/tex] .... (2)
Where, a is constant and (h,k) is vertex.
On comparing (1) and (2), we get
[tex]a=1[/tex]
[tex]h=1[/tex]
[tex]k=1[/tex]
Since the value of a is positive, therefore it is an upward parabola. The vertex of the parabola is (1,1).
Put x=0 in the given equation.
[tex]y=(0-1)^2+1[/tex]
[tex]y=1+1=2[/tex]
Therefore the y-intercept is (0,2) and option A is correct.
