Question

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.

Answer 1

Answer:

Domain:

$[0,9.146]$

Range:

$[0,12.95]$

Step-by-step explanation:

we are given

a baseball follows the path graphed by the quadratic function

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

where

d is the horizontal distance the ball traveled in yards

ƒ(d) is the height, in yards, of the ball at d horizontal yards

Domain:

we know that domain is all possible values of x for which any function is defined

So, for finding domain , we will take smallest x-value to largest x-value

so, we can set f(d)=0 and find zeros

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

we can use quadratic formula

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

$d=\frac{-54\pm \sqrt{54^2-4\left(-6\right)8}}{2\left(-6\right)}$

$d=-\frac{\sqrt{777}-27}{6},\:d=\frac{27+\sqrt{777}}{6}$

$d=-0.14579,d=9.146$

we know that d is a horizontal distance

and distance can never be negative

so, domain will be

$[0,9.146]$

Range:

Since, this is quadratic equation

so, it is also equation of parabola

so, firstly we will find vertex of parabola

Suppose, we have

$ax^2+bx+c=0$

Vertex is

$x=\frac{-b}{2a}$

so, we can compare

a=-0.6

b=5.4

c=0.8

now, we can find vertex

$x=\frac{-5.4}{2\times -0.6}$

$d=4.5$

now, we can find y-value

$f(4.5)=-0.6(4.5)^2+5.4(4.5)+0.8$

$f(4.5)=12.95$

we can see

that leading coefficient is -0.6

which is negative

so, parabola opens downward

so, range will be

$[0,12.95]$

Answer 2

Answer:

Domain is all real numbers

Range is

$y\le 12.95$

Step-by-step explanation:

The given function is

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

This is a maximum quadratic function therefore the domain is all real numbers.

Let us complete the square to find the vertex.

$f(d)=-0.6(d^2-9d)+0.8$

$f(d)=-0.6(d^2+9d)+-0.6(\frac{9}{2})^2- -0.6(\frac{9}{2})^2+ 0.8$

$f(d)=-0.6(d-\frac{9}{2})^2+\frac{243}{20}+ 0.8$

$f(d)=-0.6(d-\frac{9}{2})^2+\frac{259}{20}$

Therefore the range is

$y\le \frac{259}{20}$

$y\le 12.95$

See graph