IXL - Graph continuous piecewise-defined functions (Algebra 1 practice)

Graph this function.

(

)

– 2 x – 5	if x < – 4
1 2 x + 5	if x ≥ – 4

Select points on the graph to plot them. Select "Point fill" to change a point from closed to open.

Segment

Ray

Point fill

Edit

Segment

Ray

Point fill

Edit

ref_doc_title.

Learn with an example

question

Graph this function.

(

)

– 5 x + 5	if x ≤ 1
5 x – 5	if x > 1

Select points on the graph to plot them. Select "Point fill" to change a point from closed to open.

Segment

Ray

Point fill

Edit

key idea

A piecewise-defined function is written using two or more expressions. Each expression defines the function on a different piece of the function's domain.

An open circle at a point on the graph of a piecewise-defined function indicates the point is not included in the graph. Open circles correspond to the inequality symbols < and >.

A closed circle indicates the point is included in the graph. Closed circles correspond to the inequality symbols ≤ and ≥.

solution

Look at the piecewise-defined function:

(

)

– 5 x + 5	if x ≤ 1
5 x – 5	if x > 1

One piece is defined by the expression

–

on the interval

≤

This piece is the part of the line

–

to the left of

Evaluate the equation at

to find the right endpoint of this piece of the graph.

–

Plug in

–

(

)

Plug in

–

So, the right endpoint is

(

)

You can use the slope of

–

to find another point on this piece of the graph. Since the slope is

–

start at

(

)

and count up 5 and left 1 to find another point. The resulting point is

(

)

Plot the endpoint

(

)

and draw a ray through the point

(

)

Use a closed circle at

(

)

because

is included in the interval

≤

(

)

(

)

The other piece is defined by the expression

–

on the interval

This piece is the part of the line

–

to the right of

Evaluate the equation at

to find the left endpoint of this piece of the graph.

–

Plug in

(

)

–

Plug in

–

So, the left endpoint is

(

)

You can use the slope of

–

to find another point on this piece of the graph. Since the slope is

start at

(

)

and count up 5 and right 1 to find another point. The resulting point is

(

)

Plot the endpoint

(

)

and draw a ray through the point

(

)

Notice that

(

)

was the endpoint of the first ray. Since

was included in the interval

≤

the point

(

)

is still included in the graph

(

)

So keep the closed circle at

(

)

(

)

(

)

Looking for more? Try these:

Lesson: Piecewise functions

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Teacher tools

Work it out

Not feeling ready yet? These can help:

Q.1Evaluate piecewise-defined functions

L.4Slope-intercept form: graph an equation

L.13Graph a horizontal or vertical line

Lesson: Piecewise functions

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