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Solving
a linear-quadratic system of equations graphically
involves following a series of steps. |
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Solve
the following system of equations graphically:
y =
x2
- 4x - 2
(quadratic equation of
form
y =
ax2
+ bx
+ c
)
y = x -
2 (linear
equation of form y = mx + b) |
Step 1:
Graph one of the equations.
Let's
graph the quadratic equation first. By its form,
y =
x2-
4x - 2, we know it is a parabola.
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Rather
than picking numbers at random to form our table of values,
let's find the axis of symmetry where the turning point of the
parabola will occur. |
To
find the axis of symmetry, 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
axis of symmetry
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Since the x-coordinate
of the turning point is 2, let's use this value as the
middle
value for x in our table. We will also include 3 values above
and below 2 in our table.
Substitute
each value of x into the quadratic equation to find the corresponding values
for y and complete the table.
For
example, substituting -1 for x we get
y = (-1)2 - 4(-1) - 2
= 1 + 4 - 2 = 3
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Set up the table. |
|
x |
y |
|
-1
0
1
2
3
4
5 |
3
-2
-5
-6
-5
-2
3 |
Complete the table. |
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Next, graph
the points from the table to get the graph of the parabola at the right.
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Step one done! |
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Step 2:
Graph the second equation:
Now
graph the linear equation, a straight line, y = x - 2 on the same
set of axes. |
To
graph the straight line we need to know the slope and the y-intercept.
Remember, from the form,
y =
mx
+
b,
m
is the slope and
b is the
y-intercept. For our
equation, m = 1, b = -2. |
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Draw
the graph of the line 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 |
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Step 3:
Find the intersection points (where they cross).
The
last step is to find the point(s) where the two graphs
intersect. This is the solution set, the answer, of the system of
equations. |
Our
graphs intersect at 2 points whose coordinates are
(0,-2) and (5,3).
The solution set for this problem is:
{(0,-2),(5,3)} |
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See how to use
your
TI-83+/84+ graphing calculator with
linear
quadratic systems.
Click calculator. |
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