TI-83An Introduction to Graphing

Mathematics Staff DevelopmentLincoln Public SchoolsAugust 25, 2005© Jerel L. Welker

http://jwelker.lps.org/lessons/

Use the TRACE key to move the cursor along the line with the left and right arrow keys.

Press Y=

Graphing a FunctionGraph the function y = x - 5

Enter X - 5 in Y1=

Press ZOOM – 6 or ZStandard to set the window to set the window to [-10,10] x [-10,10]

Using the TableValues from the function can also be

viewed using the TABLE.

Use the up/down arrows to scroll through the table of values.

Press 2nd – TBLSET to change the

starting value or increment ( Tbl).

Press 2nd – TABLE to view the values in table format.

Press Y=

Enter (1/3)X + 9 in Y2

Graph the functions.

Graph a 2nd function.Graph y=x - 5 and

on the same graph.

19

3y x

To see more of the graph change:Xmin to change the left side.Xmax to change the right side.Xscl changes the increment of the tick marks on the x-axis.Ymin to change the bottom.Ymax to change the top.Yscl changes the increment of the tick marks on the y-axis.

Press WINDOW.

Changing the WindowChange the window to view the

intersection of the two functions.

Graph the functions and adjust as required.

Use the TRACE button and approximate the solution.

Method 2:

Finding the IntersectionMethod 1:

Use the TABLE (2nd - Table) to find the x-value which has the same y-value for both functions.

Press ENTER on the first curve. Use the up/down arrows to changes curves if required.

Use the intersection feature of the calculator.

Finding the IntersectionMethod 3:

Press 2nd – CALC followed by intersect or 5.

Press ENTER on the second curve.

Use the arrow keys to position the cursor close to the intersection and press ENTER. The answer will be displayed.

Clear the previous equations using the CLEAR key.

On Your Own

Find the intersection(s) of:y = -(x + 3)2 – 13y = 5x - 10

Notes:Use ^ 2 or the x2 key for a square.Note the difference between (-) and minus – keys.

On Your Own Solutions

A Complete Graph

A complete graph should:

show the origin.

show all x- and y-intercepts if they exist.

show all turning points of the graph.

show the end behavior of the graph.

utilize the entire screen.

It may be possible that some graphs require more than one screen for a complete graph!

Find a Complete GraphFind a complete graph of

y = x3 – 14x2 – 21x + 90

Hint: Only make changes to the graph in one direction at a time.

Sketch the GraphSketch the graph and show the window

in [Xmin, Xmax] x [Ymin, Ymax] form.

[-10, 18] x [-600, 150]

Showing the window provides documentation that a complete graph has been found.

Y-interceptFind the y-intercept of the function.

The y-intercept occurs where the x-value is 0. Substitute 0 for x or

Calculate the value of 0. 2nd – CALC and choose VALUE or option 1.

Enter the value of 0.

The y-intercept is (0, 90).

X-intercept(s)Find the x-intercept(s) of the function.

The x-intercepts occur on the x-axis where the y-value is 0. They are also called roots or zeros of the function.

Finding X-intercept(s)Press 2nd – CALC and choose zero or

option 2.The calculator is asking for a left bound.

Use the arrow keys to position the cursor on the LEFT side of the zero. Press ENTER.

The calculator is asking for a right bound. Use the arrow keys to position the cursor on the RIGHT side of the zero. Press ENTER.

Position the cursor over the zero and press ENTER. The zero value will be displayed.

Writing Factors

The x-intercepts or zeros are:

(-3, 0), (2, 0), and (15, 0)

Write the factors of the function.

If a is a zero of the function then:x = ax – a = 0(x – a) is a factor of the function.

x3 - 14x2 - 21x + 90 = (x + 3)(x - 2)(x - 15)

Turning PointsTurning points or extrema of the graph

are points where the slope of the line tangent to the curve is zero.

Local maximum

Local minimum

Finding a MaximumPress 2nd – CALC and choose maximum

or option 4.The calculator is asking for a left bound.

Use the arrow keys to position the cursor on the LEFT side of the maximum. Press ENTER.

The calculator is asking for a right bound. Use the arrow keys to position the cursor on the RIGHT side of the maximum. Press ENTER.

Position the cursor over the maximum and press ENTER. The maximum value will be displayed.

Finding a Maximum

The local maximum is 97.50 when x is -0.70.

Finding a Minimum

Note that the trace values take up a portion of the lower screen. If you cannot see the minimum value, reset the window to increase the space before continuing.

Finding a MinimumPress 2nd – CALC and choose minimum

or option 3.

Repeat the process of finding the LEFT, RIGHT and GUESS values for the minimum value.

The local minimum is -520.02 when x is 10.03.

DomainFind the domain of the function.

The graph is a continuous function with x starting at and continuing to .

D: ( , )

RangeFind the range of the function.

The graph is a continuous function with y starting at and continuing to .

R: ( , )

Check Your SkillsFor the function:

y = x4 + 21x3 + 123x2 + 119x - 264

Find:a complete graph and sketch.the intercepts.a linear factorization of the polynomial.the extrema or turning points.the domain and range.

Solutions - Graph

[-13, 3] x [-400, 500]

Now, find the intercepts.

For the function:y = x4 + 21x3 + 123x2 + 119x - 264

Solutions – Y-interceptFor the function:

y = x4 + 21x3 + 123x2 + 119x - 264

The y-intercept is (0, -264).

Now, find the x-intercepts.

Solutions – Zeros

Zeros are (-11, 0), (-8, 0), (-3, 0), and (1,0).

Linear Factorization (x + 11)(x + 8)(x + 3)(x – 1)

Now, find the extrema or turning points.

For the function:y = x4 + 21x3 + 123x2 + 119x - 264

Solutions – Extrema

Local min at -158.74 when x = -9.76.Local max at 223.67 when x = -5.43.Local min at -295.66 when x = -0.56.

Next, find the domain.

For the function:y = x4 + 21x3 + 123x2 + 119x - 264

Solutions – Domain

Next, find the range.

The graph is a continuous function with x starting at and continuing to .

D: ( , )

For the function:y = x4 + 21x3 + 123x2 + 119x - 264

Solutions – Range

The lowest point on the graph is -295.66 and the graph continues upward to .

R: ( 295.66, )

For the function:y = x4 + 21x3 + 123x2 + 119x - 264

Hidden BehaviorSome functions have behavior which is “hidden” by the calculator in some windows. This behavior may be extrema or discontinuities which do not show in a typical view.

For example, graph y = x3 – 0.1x in standard view (Zoom – 6).

Hidden Behavior“Flat spots” in a graph or areas hidden by the axes can be difficult spots. Use Zoom – Box or option 1 to take a closer look at these places.

Zoom Box allows the user to draw a rectangle around a portion of the graph which should be enlarged.

Hidden BehaviorPress Zoom – Box and move the cursor to a corner of the rectangle to zoom.

Hidden BehaviorPress Enter and move the cursor to the diagonal corner to form a rectangle.

Press Enter and the region within the rectangle will be shown in a new window.

Hidden BehaviorRepeat the process until a satisfactory graph is shown.

What first appeared to be a single zero is actually three zeros.

Rational FunctionsGraph:

Graph in a standard window Zoom – 6.

3( )

2

xf x

x

The numerator and denominator of rational functions must be enclosed in parenthesis. Enter the function as: Y1 = (x + 3)/(x – 2)

Y-InterceptFind the y-intercept.

Calculate the value of f(x) when x=0.

30,2

The y-intercept is

Rational FunctionsGraph:

3( )

2

xf x

x

In connected mode, some may be tempted to say the asymptote shows on the graph.

Rational Functions

Change the mode to DOT by pressing MODE and pressing ENTER with the cursor on Dot.

Graph the function.

Rational FunctionsIn Dot mode, only points graphed by the grapher are displayed. No connecting lines are shown. Note the “line” is not shown in this view and is not part of the graph.

If the line were part of the graph, there must be an x-intercept where the line crosses the x-axis. Solve for the x-intercepts algebraically.

X-Intercept3

( )2

xf x

x

0 3x

30

2

x

x

3x

(x-2)(x-2)

The only zero is x = -3 which is shown on the graph. There is no zero near the “line” on the connected graph.

X-InterceptCalculate the function value when x = -3.

The grapher confirms the function value of -3 is zero.

DomainThe function is discontinuous. It cannot be traced without lifting a pencil near x=2.

What happens as x = 2? What does a blank Y= result indicate?

Domain

x f(x)

1.9 -49

1.99 -499

1.999 -4,999

1.9999 -49,999

1.99999 -499,999

1.999999 -4,999,999

What happens as x approaches 2? Use the calculate value function to investigate. Calculate the values of f(x) as x approaches 2.

x f(x)

2.1 51

2.01 501

2.001 5,001

2.0001 50,001

2.00001 500,001

2.000001 5,000,001

2

3lim

2x

x

x

2

3lim

2x

x

x

DomainWhat is the domain of the function?

The function is continuous at all points except for x = 2.

( , 2) (2, ) D:

The function has a vertical asymptote at x = 2.

RangeWhat is the range of the function?

We have already discovered that the graph goes to and as x approaches 2. What happens as the graph extends horizontally?

Extend the window to show more “x-values”. Trace toward the ends to find what happens to the function values as x approaches .

RangeTracing to both ends of the graph, we find that the function value approaches what value?

3lim 1

2x

x

x

3

lim 12x

x

x

RangeWhat is the range of the function?

The function begins at and increases until approaching the horizontal asymptote at 1. From 1, the graph continues to .

( ,1) (1, ) R:

The function has a horizontal asymptote at y = 1.

Check for Understanding

Given g(x), find:

4 2( )

3

xg x

x

• the intercepts.

• the domain and range.

• the asymptotes.

Check for Understanding

Given g(x), find:

4 2( )

3

xg x

x

• the intercepts.

• the domain and range.D: R:

• the asymptotes.y = -2 and x = -3

40, 2, 03

and

( , 3) ( 3, ) U ( , 2) ( 2, ) U

Hiding a Graph

3 6( )

2

xh x

x

Enter in Y2.

At times, one may wish to keep a graph for later use but not be visible in the grapher at all times. It is possible to “hide” the graph from view without clearing the equation from the grapher.

Leave the current equation in Y1.

Hiding a GraphHide the graph in Y1.

Position the cursor over the = sign in the equation line and press ENTER.

Move the cursor away from the = sign.

When the = box is highlighted, the function will graph.

When the = box is not highlighted, the function will not graph.

Repeating the process of pressing ENTER on the = sign will highlight the box and graph the function.

Hiding a Graph

Both Y1 and Y2 will graph.

Only Y2 will be visible.

Hidden Behavior3 6

( )2

xh x

x

View the graph of

Is the graph what you expected?

Should there be an asymptote at x = 2?

What is the value of f(2)?

Hidden Behavior

Start with a window with Xmin of -10 and Xmax of 10. Change the window so that x=2 is in the center of the window, a “hole” appears in the graph.

This “hole” in the graph is another example of hidden behavior.

D:

D: or

Domain and Range

The domain is all real numbers except x = 2. This could be written as:

, 2 2, U

| 2x x

The range is | 3y y

Removable DiscontinuityThis particular “hole” in the graph is called a removable discontinuity. By redefining one point, the graph is a continuous function.

3 6 3( 2)( ) 3

2 ( 2)

x xh x

x x

Simplifying the function h(x), we get:

Since the “hole” appears where x=2 and h(2) = 3, the graph has a removable discontinuity at (2, 3). Graphing h(x) and graphing the point (2, 3) would result in a continuous function.

Removable Discontinuity

A removable discontinuity occurs at the zero of a common factor of the numerator and denominator of a rational function. If x = c is a zero of the common factor, the point (x, f(c)) is a removable discontinuity.

Find the removable discontinuity of:2

2

4( )

3 2

xm x

x x

Removable DiscontinuityFind the removable discontinuity of:

2

2

4( )

3 2

xm x

x x

2

2

4 ( 2)( 2) 2( )

3 2 ( 2)( 1) 1

x x x xm x

x x x x x

Since (x + 2) is a common factor of the numerator and denominator, a removable discontinuity occurs at x = -2

If the common factors are removed, m(-2) is 4. The removable discontinuity is at (-2, 4).

Removable DiscontinuityRemovable discontinuity at (-2, 4)

The line connecting the top dot with the bottom dot is not a part of the graph. The graph should have a vertical asymptote at x = -1.

Try calculating the values of m(-2) and m(-1). What are they?