Chapter 9.2, Problem 27PPS

(a)

To determine

the value of g(x) that is obtained from the graph of f(x) = x²

Answer to Problem 27PPS

The graph of g(x) is the graph of f (x) translated 200 yards right, compressed vertically, reflected in the x-axis, and translated up 20 yards.

Explanation of Solution

Given information:

The path of a drive can be modeled by a quadratic function where g(x) is the vertical distance in yards of the ball from the ground and x is the horizontal distance in yards.

  

Formula used:

Substitution method is used.

Calculation:

For

  $\begin{array}{l}g\left(x\right)=0.0005{\left(x–200\right)}^2+20,\\ a=–0.0005,\\ h=200,and\\ k=20.\\ \text{Since }\\ \left|–0.0005\right|<1\text{ and }\\ –0.0005<0,\end{array}$

There is a reflection across the x-axis and the graph will be compressed vertically (or be wider). 

Since

  $h=200\text{ and }200>0$

There is a translation 200 yards to the right.

Since

k = 20 and 20 > 0, there is a translation up 20 yards.

Thus, the graph of g(x) is the graph of f (x) translated 200 yards right, compressed vertically, reflected in the x-axis, and translated up 20 yards.

Conclusion:

The graph of g(x) is the graph of f (x) translated 200 yards right, compressed vertically, reflected in the x-axis, and translated up 20 yards.

(b)

To determine

the function h(x) to determine the second golfer shot.

Answer to Problem 27PPS

The function is $h\left(x\right)=0.0005{\left(x–230\right)}^2+20$

Explanation of Solution

Given information:

GOLF The path of a drive can be modeled by a quadratic function where g(x) is the vertical distance in yards of the ball from the ground and x is the horizontal distance in yards.

  

Formula used:

Substitution method is used.

Calculation:

 If the red tee is 30 yards closer to the hole, then g (x) will be translated 30 yards to the right to obtain h (x).

So, for h(x) h = 200 + 30 or 230. Therefore, the function is $h\left(x\right)=0.0005{\left(x–230\right)}^2+20$

Conclusion:

The function is $h\left(x\right)=0.0005{\left(x–230\right)}^2+20$