We know that the bottom of the river can be modeled with the given absolute value function.

d (h) = \frac{1}{5} ∣ h - 240 ∣ - 48

In the above equation,

d

is the depth of the river bottom (in feet), and

h

is the horizontal distance to the left-hand shore (in feet). Since the buoys will be placed where the river bottom is

30

feet below the surface, to know how far from the left-hand shore we should place them we will substitute

- 30

for

d (h)

in the given equation.

- 30 = \frac{1}{5} ∣ h - 240 ∣ - 48

We will solve the above equation graphically. To do so, we will start by by isolating the absolute value expression.

- 30 = \frac{1}{5} ∣ h - 240 ∣ - 48

AddEqn

$LHS + 48 = RHS + 48$

18 = \frac{1}{5} ∣ h - 240 ∣

MultEqn

$LHS \cdot 5 = RHS \cdot 5$

90 = ∣ h - 240 ∣

RearrangeEqn

Rearrange equation

∣ h - 240 ∣ = 90

We can think of the left-hand side of our simplified equation as the absolute value function

y = ∣ h - 240 ∣ .

If we graph this function, the solutions to our equation will be the

x -

coordinates of the points on the graph whose

y -

coordinates are

90 .

The $x -$ coordinates of the points on the graph whose $y -$ coordinates are $90,$ are $150$ and $330 .$ Therefore, the solutions for our equation are $x = 150$ and $x = 330 .$ This means we should place the buoys at $150$ feet and at $330$ feet from the left-hand shore.

Exercise