Lesson Page

Math A

Solving a Linear-Quadratic System Graphically

 

Solving a linear-quadratic system of equations graphically involves following a series of steps.

The easiest way to illustrate this is to follow the steps through in a specific example-------so, here goes.

Example 1
1. Solve the following system of equations graphically:
         y = x2 - 4x - 2
         y = x - 2          
First we will graph the quadratic equation:  
          y = x2- 4x- 2
We recognize this as the graph of a parabola, since it fits the form:
     y =
ax2 + bx + c

 

Since no values are specified for x, we will first find the equation of the axis of symmetry.

To do this we use the formula
          x = -b/2a
In this example, a =1, b=-4,
       and c = -2.

Substituting we get:
        x = -(-4)/2(1)
        x = 4/2
        x = 2

 

Now we know that the turning point, which lies on the axis of symmetry, has an x-coordinate of 2.

 

This will be our middle value for x.  We will choose 3 values less than 2 and 3 values greater than 2.

 

 

 

 

x

y

-1
0
1
2
3
4
5

 

 

 

 

 

Substitute each value of x in the quadratic equation to find the corresponding values for y. For example, substituting -1 for x we get
y = (-1)2 - 4(-1) - 2
y = 1 + 4 - 2
y = 3

 

If we do this for each value of x we get the table to the right.

 

 

 

 

x

y

-1
0
1
2
3
4
5

3
-2
-5
-6
-5
-2
3

 

Next we graph the points from the table to get the graph of the parabola to the right.

 

 

 

 

 

 

 

O.K.

Done with
the first
step !

 

Now we'll graph the linear equation  y = x - 2  on the same set of axes. To do that we need to know the slope and the y-intercept.
Standard form for a line is:
      y = mx + b
where m is the slope, and b is the y-intercept.
Our equation is y = x - 2, so
    m = 1, and b = -2
Draw the graph starting at  -2 on the y-axis.  Use slope (which is rise over run) to find other points by going up 1 and to the right 1, or down 1 and to the left 1.


y = x2 - 4x - 2
y = x - 2

 

The last step is to find the point(s) where the two grpahs intersect.  This is the solution set of the system of equations. Our graphs intersect at 2 points whose coordinates are:
(0,-2)  and  (5,3)

So the solution set is:
{(0,-2),(5,3)}

 

 

 

How to use your
TI-83+ graphing calculator  with quadratic equations.
Click calculator.

 

 


   Murray