Linear-Quadratic Systems
(Line - Parabola, Line - Circle)
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The graphing capabilities of the calculator will be put to use to solve linear-quadratic systems.
This solution method will simply graph the equations and find the points of intersections,
if such points exist.
Line and Parabola:
A quadratic equation is defined as
an equation of degree two.
In a linear- quadratic system where only one variable in the quadratic is squared,
the graphs will be a parabola and a straight line.
Linear - quadratic system - Line and Parabola:
(where only one variable is squared) |
When graphing a parabola and a straight line on the same set of axes, three situations are possible.
 |
 |
 |
| The equations
will intersect in two locations. Two real solutions. |
The equations
will intersect in one location. One real solution. |
The equations will not
intersect.
No real solutions. |
| Solve graphically: |
y = -x2 + 2x + 4 (quadratic -
parabola)
x + y = 4 (linear) |
| 1. |
• If needed rewrite the equation so they are in "y =" form.
• In this example, change the linear equation
to "y=" form. |
y = -x + 4 |
| 2. |
Enter
the equations as "f 1(x)" and
"f 2(x)".
Be sure you can see the intersection points on the screen. Adjust the window if needed.
|
|
| 3. |
Use the Intersection option.
 , #6 Analyze Graph,
#4 Intersection
Repeat for each intersection point.
Solution: (0,4) and (3,1) |
 |
Line and Circle:
In a linear- quadratic system where BOTH variables in the quadratic are squared,
the graphs will be a circle and a straight line.
Linear - quadratic system - Line and Circle:
(where both
variables are squared) |
Like with the parabola, the graphs below show that when a line and a circle
are graphed on the same set of axes, three situations are possible.
 |
 |
 |
| The equations will intersect in two locations.
Two real solutions. |
The equations will intersect in one location (a tangent). One real solution. |
The equations will not intersect.
No real solutions. |
| Solve graphically: |
x2 +
y2 = 25 (quadratic - circle)
x - y = 5 (linear) |
|
| 1. |
• When graphing a line and a circle on the same grid, it will be easier if the line is graphed first.
•
Rewrite the linear equation into "y =" form.
•
Enter
the equation as "f 1(x)". |
Graph the "line" first!
y = x - 5 |
| 2. |
We know that this circle has a center at the origin, (0,0), and a radius of 5.
(h,k) = (0,0) and r = 5
, #3 Graph Entry/Edit, #3 Equations Template, #3 Circle, #1 Center form
Adjust window if needed to see intersection points. |
|
| 3. |
Use the Intersection option.
, #6 Analyze Graph,
#4 Intersection
In this situation, the "intersection" option will create a "box" around the point.
Click to lock a corner of the box.
Drag to form the box.
Repeat for each intersection point. |
 |
| |
Solution: (0,-5) and (5,0) |
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