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The point $(8,2.4)$ is on the graph. Using these three points, two function rules can be written. One for the decreasing part, and the other for the increasing part of the graph. The domain of these parts can be written as follows. Note the point at $x=4$ can belong to either piece, so long as it belongs to only one of them. The next step will then be to find the equation for both lines. For this, recall the slope-intercept form of a linear function. In this form $m$ is the slope of the line and $b$ is the $y\text{-}$intercept. Start with the decreasing part. The $y\text{-}$intercept and slope can be found from the graph. For this line the $y\text{-}$intercept is $b=2.4$ and the slope is $m=\text{-}0.6.$ Similarly, an equation for the increasing part can also be written. For the increasing line, the $y\text{-}$intercept is $\text{-}2.4$ and the slope is $0.6.$ Knowing the equations for both lines, the absolute value function can be written as a piecewise function. To do so, the function rules can be rearranged so that one rule contains an expression and the other rule contains its opposite. The function rules contain the expressions $\text{-}(x-4)$ and $x-4.$ These expressions produce non-negative values in their domains. Therefore, they can be written using absolute values. The domain of this absolute value function is the union of the domains for the function rules.

The function rules can be simplified. Next, each function rule will be drawn separately and their graphs will be combined. First draw the graph of $y=0.6x.$ The graph ends with an open circle because its domain is the set of $x\text{-}$values less than $200.$ For $x\text{-}$values greater than or equal to $200,$ the graph of $y=\text{-}0.6x+240$ will be drawn. This is a linear function written in slope-intercept form. Using this information, its graph can be drawn. This piece ends with a closed circle, as $200$ is in its domain. Finally, both graphs will be combined on the same coordinate plane.

Since all the measures are in meters, the distance between these two points is $400$ meters.

Considering the definition of absolute values, this inequality can be divided into two pieces. The expressions can be simplified. Now, the inequalities describing the domains of the pieces will be rearranged. To do so, $1.5$ will be added to both sides of the inequalities. The boundary line is in slope-intercept form. Its graph can be drawn by using the slope and $y\text{-}$intercept. Since the inequality is not strict, the boundary line will be solid. Additionally, since $x$ is less than $1.5,$ the boundary line will have an open endpoint at $x=1.5.$ To determine the region to be shaded, choose a test point and substitute it into the inequality — for example, $(0,0).$ Since the point does not satisfy the inequality, the region that does not contain the test point will be shaded. Note that only points in the corresponding domain $x<1.5$ will be considered. Next, the second piece of the inequality will be graphed. The boundary line's slope is $5.$ Also, $(1.5,\text{-}1.5)$ satisfies the second piece. This point is the leftmost point of the boundary line because its domain is $x\ge 1.5.$ Additionally, since the inequality is non-strict, the boundary line will be solid. Now, choose a point not on the boundary line whose $x\text{-}$coordinate is greater than or equal to $1.5$ — for example, $(4,0).$ Since the point did not satisfy the inequality, the region that does not contain the point that will be shaded. Finally, the graphs of the two pieces can be combined on the same coordinate plane. Maya can now complete her image.

The absolute value expression in this function can be divided into two cases.

• If $t-25$ is less than $0,$ then its absolute value is equal to its opposite value.
• If $t-25$ is greater than or equal to $0,$ then its absolute value is equal to itself.

With this information in mind, the absolute value function can be written as a piecewise function. Next, the function rules need to be simplified. The inequalities can be solved for $t.$ To do so, $25$ will be added to both sides of the inequalities. First graph $y=\text{-}2t+50.$ This function has a slope of $\text{-}2$ and a $y\text{-}$intercept of $50.$ Since the domain for this piece is the set of numbers less than $25,$ the graph should end with an open point at $t=25.$

Similarly, the other piece $y=2t-50$ can be drawn. Its slope is $2.$ The domain of this piece contains the $t\text{-}$values greater than or equal to $25.$ Therefore, it has a closed endpoint at $t=25.$

The combination of the above graphs is the graph of the piecewise function.

Since $t$ represents time and $d(t)$ represents distance, they cannot be negative. Therefore, both $t$ and $d(t)$ are non-negative numbers. Recall that Jordan swims to the far end of the pool and comes back to the starting point. At $t=0,$ Jordan is at the starting point. Therefore, by substituting $t=0$ into the given function the length of the pool can be found. At $t=0,$ Jordan is $50$ meters away from the end. With this information it can be said that $d(t)\le 50.$ When Jordan is back to the starting point, she is again $50$ meters away from the far end. Therefore, to calculate how many seconds it takes her to swim to the far end and come back to the starting point, $d(t)=50$ will be substituted in the given function. Jordan is $50$ meters away from the far end at $t=0$ and $t=50.$ Therefore, it takes her $50$ seconds to swim to the far end and come back to the starting point. With this information it can be said that $t\le 50,$ so the domain and the range can be written. This means that only the first quadrant is relevant.