Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
End-of-Course Assessment

Exercise 15 Page 965

Take notice of where the closed and open circles are.

A

Practice makes perfect

To write the piecewise function for the given graph, we need to find the equation of the line representing each piece and then restrict the domain accordingly.

The First Piece

Let's begin by finding the slope and y-intercept of the first piece of the function. To help us write the equation of the line for this piece, we can extend the line to the y-axis. This will help us to know the y-intercept.

Since this piece of the function is a horizontal line, it has a slope of 0. We can also see that the y-intercept is 2. Let's write the equation for this piece in slope-intercept form. y= 0x+ 2 ⇔ y=2

The Second Piece

Now let's take a look at the second piece.

The line has a slope of 1 and the y-intercept is 3. Let's write the equation for this piece in slope-intercept form. y= 1x+ 3 ⇔ y=x+3

The Third Piece

Finally, let's take a look at the third piece. To help us write the equation of the line for this piece, we can extend the line to the y-axis. This will help us to know the y-intercept.

We can see that this line has a slope of - 1 and the y-intercept is 1. We will write the equation for this piece in slope-intercept form. y= - 1x+ 1 ⇔ y=- x+1

Combining the Pieces

We can add the equations of these lines to the piecewise function notation. f(x)= 2 x+3 - x +1 Finally, we need to determine the domain for each equation. The jump occurs at x=- 1 and x=2, so this will be where the domains are divided. We cannot have an overlap in our domains so we need to take notice of where the closed and open circles are located.

Since we cannot have an overlap, we will assume that the first piece includes -4 but will not include - 1. The closed circle at the second piece tell us that its domain includes - 1 and the open circle means that 2 is not included. The closed circles at the third piece tells us that its domain includes 2 and 4. f(x)= 2, &- 4≤ x < - 1 x+3, &- 1 ≤ x < 2 - x +1, &2≤ x ≤ 4 This corresponds to option A.