Graphing Absolute Value Equations and Inequalities as Systems

We need to write the given absolute value equation as a piecewise function. y = - 4|x-7| We can rewrite the expression |x-7| by using the definition of an absolute value. If x-7 is less than 0, its absolute value is equal to its opposite value. Conversely, if x-7 is greater than or equal to 0, its absolute value is equal to itself. |x-7| = - (x-7) & if x-7 < 0 x-7 & if x-7 ≥ 0 Considering this definition, we can rewrite the given function. y = - 4 [ - (x-7)] & if x-7 < 0 - 4 ( x-7) & if x-7 ≥ 0 Now we simplify each function rule.

Finally, we solve the inequalities describing the domain of each piece. To do so, we need to add 7 to both sides of the inequalities. y= 4x-28 & if x < 7 - 4x+28 & if x ≥ 7 This corresponds to option A.

Again, we need to write the absolute value function y= 5|x-8|+3 as a piecewise function. Let's start by recalling how to write the vertex form of an absolute value function, g(x)= a|x- h|+ k, as a piecewise function. g(x) = a[-(x- h)]+ k, if x- h < 0 a(x- h)+ k, if x- h ≥ 0 Using this, let's identify the values for our function. g(x) = & a|x- h|+ k & ⇓ y = & 5|x- 8|+ 3 Now, we will substitute a= 5, h= 8, and k= 3 into the above piecewise function.

We need to solve the inequalities describing the domain of each piece. y= - 5x+43, & if x< 8 5x-37, & if x ≥ 8 This corresponds to option C.

We need to write the given absolute value equation as a piecewise function. y = |3x+6|+4 We can rewrite the expression |3x+6| by using the definition of an absolute value. If 3x+6 is less than 0, its absolute value is equal to its opposite value. Conversely, if 3x+6 is greater than or equal to 0, its absolute value is equal to itself. |3x+6| = - (3x+6) & if 3x+6 < 0 3x+6 & if 3x+6 ≥ 0 Considering this definition, we can rewrite the given equation.

Finally, we solve the inequalities describing the domain of each piece. To do so, we need to subtract 6 from both sides and then divide each side by 3. y= - 3x-2 & if x < - 2 3x+10 & if x ≥ - 2 The resulting piecewise function corresponds to option B.

Let's use the piecewise function found in the previous part to graph the given function. We will draw each function rule separately and then combine them. y= - 3x-2 & if x < - 2 3x+10 & if x ≥ - 2 First, we will draw the graph of y=- 3x-2. y = - 3x - 2 ⇒ l Slope: - 3 y-intercept: - 2 Note that the domain for this piece is x < - 2. Let's draw it.

The graph ends with an open circle because its domain is the set of x-values less than - 2. For x-values greater than or equal to - 2, we will draw the graph of y=3x+10. y = 3x + 10 ⇒ l Slope: 3 y-intercept: 10 Let's draw it.

This piece starts with a closed circle, as - 2 is in its domain. Finally, we combine the graphs on the same coordinate plane.

Therefore, the answer is option A.

Let's first write the given absolute value function as a piecewise function. Then, we will identify the error in the presented work. y = - |- 1/3x+1 |-1 We can rewrite the expression | - 13x+1 |. If - 13x+1 is less than 0, its absolute value is equal to its opposite value. Conversely, if - 13x+1 is greater than or equal to 0, its absolute value is equal to itself. |- 1/3x+1 |= - (- 13x+1 ) & if - 13x+1 < 0 [0.5em] - 13x+1 & if - 13x+1 ≥ 0 Considering this definition, we can write a piecewise function.

Finally, we solve the inequalities. To do so, we need to subtract 1 from both sides of the inequalities and multiply by - 3. Note that multiplying by a negative number flip the inequality symbol. y= - 13x & if - 13x+1 < 0 [0.5em] 13x-2 & if - 13x+1 ≥ 0 ⇓ y= - 13x & if x > 3 [0.5em] 13x-2 & if x ≤ 3 Comparing the resulting function with the one Dylan wrote, we see that the inequality symbols were not flipped. Therefore, Statements I and III apply in this case. Analyzing each of those statements, let's consider the characteristics of their graphs, respectively.

We can use this information to now draw their graphs.

Notice that the graph of Piece II ends with a closed circle. Therefore, Statement II also corrects Dylan's mistake. As a result, we will choose the first three statements, I, II and III.

We are given the graph of an absolute value function.

As we can see, the behavior of the absolute value function changes at the vertex. Therefore, this is a good point to separate the graph into two pieces, both being straight lines. Our next step will be to find the equation for both lines. For this, let's recall the slope-intercept form of a line. y=mx + b In this form, m is the slope of the line and b is the y-intercept. We can start with the line to the left of the vertex first. We can identify the slope from the given graph.

For this line, the slope is m=5. We can substitute a point on the graph to find the y-intercept. Let's use (- 7,- 3).

Therefore, the equation for the left-hand part is y=5x+32. We will now do the same with the line to the right of the vertex.

Then, for this second line, the slope is m=- 5. To find the y-intercept, we can substitute (- 5,- 3).

Therefore, the equation for the right-hand part is y=- 5x-28. Knowing the equations for both lines allows us to write the absolute value function as the combination of two functions. y= 5x+32 - 5x-28 Last, let's decide the domain of each piece. The pieces meet and change directions at x=- 6, so this is where our domain needs to change. The point at x=- 6 can belong to either piece, as long as it belongs to only one of them. This means that we can correctly write the piecewise function in two different ways. y= 5x+32, & if & x ≤ - 6 - 5x-28, & if & x > - 6 or y= 5x+32, & if & x < - 6 - 5x-28, & if & x ≥ - 6 Therefore, the correct answer is option C.

We have been given a graph that shows the path of sunlight as it reflects off the water.

To write this as an absolute value function, the point of reflection is going to be the vertex. Recall the vertex form of an absolute value function. f(x)=a|x- h|+ k Here, the point ( h, k) is the vertex. Let's substitute our vertex ( 5, 0) into this equation. f(x)=a|x- 5|+ 0 ⇓ f(x)=a|x-5| Now we can solve for a by substituting another known point into the equation. Let's use ( 4, 2).

We can now write our final equation. f(x)=2|x-5| Note that the absolute value function can also be written as f(x)=|2x-10|.

To write the absolute value function as a piecewise function, we first need to determine where the function changes its direction because this is where our domains will split. In this case, this occurs when x=5.

We will let the domain of the left piece be all real numbers less than 5. Domain Left Piece: x < 5 The right piece of the function will then have a domain that includes all real numbers greater than or equal to 5. Domain Right Piece: x ≥ 5 Next, we can write the function as two separate equations by splitting our absolute value into its two cases: one positive and one negative. f(x)= 2[-(x-5)], & x < 5 2(x-5), & x ≥ 5 We can simplify these equations a little bit. Let's start with the negative case.

Now we can simplify the positive case. f(x)=2(x-5) ⇓ f(x)=2x-10 Finally, we can write the rule of the piecewise function. f(x)= -2x+10, & if x< 5 2x-10, & if x≥ 5 Therefore, the correct answer is option A.