|
A quadratic equation is defined as
an equation in which one or more of the terms is squared but
raised to no higher power.
The general form is ax2 + bx + c
= 0, where a, b and c are constants.
|
 |
|
We learned in Algebra that
linear quadratic systems can be solved algebraically and
graphically. See the Algebra site if you need practice
with algebraic solutions. In Geometry, we will concentrate
on the graphical solutions to these systems. |
Linear - quadratic system:
(where the quadratic is in one variable - only one
variable is squared) |
|
y =
x - 2
(linear)
y =
x2 - 4x - 2
(quadratic - a parabola) |
This familiar linear- quadratic system, where only
one variable is squared in the quadratic, will be the graph of a
parabola and a straight line. When a parabola and a straight
line 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. One real solution. |
The equations will not
intersect.
No real solutions. |
|
Solve
this problem manually (without graphing calculator): |
|
y =
x2 - 4x - 2
y =
x - 2
|
| 1. |
We will start with
the graph of the quadratic equation. You can start with
either equation.
|
y =
x2-
4x- 2 |
| 2. |
We want to produce a representative
graph of the parabola, one that will show the turning point. Find the equation of the axis of symmetry
to guarantee that the turning point of the parabola will be
graphed. |
|
Axis of
symmetry formula: |
 |
|
a = 1, b = -4, c = -2 |
|
Substituting: |
 |
| |
 |
| |
x = 2
|
|
| 3. |
Choose this new x
value as the center of a domain for graphing the parabola.
Create a chart of values. Three
values are usually tested above and below this x-value.
Substitute
each value of x in the quadratic equation to find the corresponding values
for y to complete the chart.
|
|
|
|
x |
y |
|
-1
0
1
2
3
4
5 |
3
-2
-5
-6
-5
-2
3 |
|
|
| 4. |
Plot these points on graph paper.
Sketch the parabola. |
 |
| 5. |
Now, graph the linear equation
on the same set of axes. You can use the same table of
values and simply find the y values for the straight
line. Or you can use the slope and y-intercept to
graph the line. |
Standard form for a line is:
y =
mx + b
where
m
is the slope, and b is the y-intercept.
For
y =
x - 2,
m = 1, and b = -2
Draw
a starting point at -2
on the y-axis. Use slope (rise over run) to find a
second point by going up
1 and
to the right
1, or down
1 and
to the left
1. |
| 6. |
Using a straight-edge, draw the
line. |
 |
| 7. |
Last, find the point(s) where the two graphs
intersect (if they do intersect). |
These
graphs intersect at 2 points whose coordinates are:
(0,-2) and (5,3)
Solution Set:
{(0,-2),(5,3)} |
| 8. |
Let's check these answers to be sure we are correct. Place
each point into BOTH equations. |
Check (0,-2)
y =
x2 - 4x - 2
-2 =
02 - 4(0) - 2
-2 = -2
check |
y =
x - 2
-2 = 0 - 2
-2 = -2 check |
Check (5,3)
y =
x2 - 4x - 2
3 =
52 - 4(5) - 2
3 = 3 check |
y =
x - 2
3 = 5 - 2
3 = 3 check |
|
| |
Wasn't it nice how the points of intersection
turned out to be nice integer values that were easily seen on
the graph! |
|
|
The problem we just solved worked out "nicely" since the
intersection points were easily seen as integer values on the
graph paper. Of course, this does not happen with all
graphs. How could we tell, by looking at the graph, if a
point of intersection was (2.3, 1.5), for example?
Answer: We could not tell
"by looking". We would have to solve the system
algebraically to find such an intersection point, or we would
have to use our graphing calculator with the intersect option.
|
|
Solve
this problem using a graphing calculator: |
|
y =
x2 + x - 1
y =
-2x + 1 |
|
Solve this system of equations
(rounding answer(s) to three
decimal places if necessary):
|
Enter
both equations into Y=. |
Choose the intersect option #5.
|
Find each point (one at a time).
|
|
If you
need additional instructions
for working with your
graphing calculator: |
|
 |
See how to use your
TI-83+/84+ graphing
calculator with
quadratic-linear systems.
Click calculator. |
|
|
Solution Set:
{(0.562,-0.123), (-3.562,8.123)} |
|
Linear - quadratic system:
(where the quadratic is in two variables - both
variables are squared) |
|
4y =
3x
(linear)
x2
+ y2 = 25
(quadratic
- a circle)
|
|
Solve
this problem manually (without graphing calculator): |
|
|
| 1. |
We will start with
the graph of the quadratic equation, the circle. You can
start with either equation.
The general form of a circle with
center at the origin is
x2
+ y2 = r2,
where r is the radius.
|
For this equation,
x2
+ y2 = 25, the center is at the origin,
(0,0) and the radius is 5 (the square root of 25). |
| 2. |
It is not necessary to make a chart to produce this graph.
Since we know the center and the radius, we can easily draw the
graph. A compass can be used to make a more accurate looking
graph. |
 |
| 3. |
Now, graph the linear equation
on the same set of axes. Using the slope and y-intercept
of the line will produce the graph quickly. Use a
straight-edge to draw the line.
|
4y =
3x
The y-intercept
is 0 and the slope is 3/4. |
| 4. |
Last, find the point(s) where the two graphs
intersect (if they do intersect). |
The graphs
intersect at points (-4,-3) and (4,3).
Solution Set:
{(-4,-3), (4,3)} |
|
