Question

1-Suppose the path of a baseball follows the path graphed by the quadratic function ƒ(d) = –0.6d2 + 5.4d + 0.8 where d is the horizontal distance the ball traveled in yards, and ƒ(d) is the height, in yards, of the ball at d horizontal yards. Identify the domain and range that matches this situation
Domain: [0,12.95] horizontal yards, range: [0,9.15] vertical yards
B. Domain: [0,9.15] horizontal yards, range: [0,12.95] vertical yards
C. Domain: [0,12.95] vertical yards, range: [0,9.15] horizontal yards
D. Domain: [0,9.15] vertical yards, range: [0,12.95] horizontal yards



2-Which of the following is the graph of y = (x – 1)2 + 1?

Answer 1

Answer:

1. The correct option B.

2.The correct option A.

Step-by-step explanation:

The given function is

$f(d)=-0.6d^2+5.4d+0.8$

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.

$0=-0.6d^2+5.4d+0.8$

Quadratic formula:

$d=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Using the quadratic formula we get

$d=\frac{-5.4\pm \sqrt{(5.4)^2-4(-0.6)(0.8)}}{2(-0.6)}$

$d=-0.146,9.146$

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.

$9.146\approx 9.15$

Therefore the correct option is B.

2.

The given equation is

$y=(x-1)^2+1$                 .... (1)

The standard form of parabola is

$y=a(x-h)^2+h$             .... (2)

Where, a is constant and (h,k) is vertex.

On comparing (1) and (2), we get

$a=1$

$h=1$

$k=1$

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.

$y=(0-1)^2+1$

$y=1+1=2$

Therefore the y-intercept is (0,2) and option A is correct.