Finding Points for Graphing a Function with a TI-83/84

Website Home                                            Mobile Pages Index

General:
This procedure is aimed at finding ordered pairs to graph a function by hand.  There are several methods for doing that with a calculator, but they require several operations that are not intuitive.  So, students often forget the procedure of the more non-intuitive methods. We will use only the graphing and TRACE functions.

Preliminaries:

We will demonstrate how to find general coordinates of points for graphing a linear equation in two variables.  We will then find the coordinates for the specialized method of the intercept method.  Finally, we will find the points for graphing a parabola.
Method for General Points:
Let's graph the equation -x+y=1.

a) First the equation must be put in the y=mx+b form so that it can be entered in the calculator.

y=x+1

b) Press Y= and enter the equation opposite Y1=; then press GRAPH.
Let's choose x-values -2, 0, 3 as indicated in the completed table below.

c) Press TRACE and then 2, the first x-value, and then press ENTER to get y=3.

d) Press 0 to get y=1 and 2 to get y=3.

That's all.  The table of values should look like the one below.

 x y -2 -1 0 1 2 3

Now graph the points and draw a straight line through them.

Method for Graphing with Intercepts:

Let's graph the equation 2x+3y=3.

a) First the equation must be put in the y=mx+b form so that it can be entered in the calculator.

y=-2/3x+1

b) Press Y= and enter the equation opposite Y1=; then press GRAPH.

c) Press ZOOM, 4 for ZDecimal to get "friendly" values of x and y.
d) Press TRACE and then 0, ENTER to get y=1.  You now have x=0 and y=1 for the y-interecept.

d) Using the left cursor arrow, move the cursor until y=0 appears at the bottom of the screen.  You will then have
x=1.5, y=0 for the x-intercept.

Now graph the points and draw a straight line through them.

Method for Graphing a Parabola:

Let’s find the points for graphing the following equation for a parabola f(x) = 2x2-4x+1.
a.  First we should find the x-coordinate for the
vertex of the parabola since that gives us a good starting place to locate additional points.  That coordinate can be found with the
following equation:  x=-b/2a   (Where “a” is the number in front of the x2 term and “b” is the number in front of x, and c is the constant term.)
Now, calculate and record the x-coordinate of the vertex.
a) From the home screen of the calculator enter -(-4)/(2*2) to get 1.  The two signs in front of 4 are made with the (-) key on the calculator.
Plug that value, 1, into the equation quadratic equation to get 2(1)2 -4(1)+1 =-1.  The points for the vertex are (1,-1).  Enter this value in
First graph the equation:
a)  Press Y=  (Located at the left end of the screen.), and clear any equations already entered by using the CLEAR button.
b)  Press, the [X,T,θ, n] button to enter x.  Then press the x2 button to get X2
c)  Press – (the minus sign), 4, X , +1.  You should now have 2X2 -4X +1 opposite Y1=.
d)  Press GRAPH to graph the equation and when the graph appears press TRACE. (Trace is one of the buttons immediately below the screen.)
Now find the points:
a)  Press ZOOM; then press 4 for ZDecimal. If the vertex of the parabola is not on the screen press ZOOM, 3 for ZoomOut.  Move the cursor until
you are on the center of the screen, x=0, y=0 and press ENTER.
b) Press TRACE and then move the cursor along the graph until you are near the vertex. Carefully move the cursor through the vertex until you
get to x=1, y=-1. Note that if you move the cursor to either side of that value, the absolute value of  y=  will be smaller.    Let’s enter these
values in a   table; then calculate more points.

 x y -1 7 0 1 1 -1 2 1 3 7

c) Enter 0 and press ENTER to get y=1; then press -1, ENTER to get 7.  Note that by symmetry, the value for x=2 is the same as for x=0 and the
value for 3 is the same as for -1.
d)  Enter the values on a coordinate system and sketch the graph through the points.

Released:  4/12/12
Last Revised: 4/7/13