Beginning Algebra
Answer/Discussion to Practice
Problems
on Intercepts
 Answer/Discussion
to 1a
2x - 3y = -6 |
 |
*Find x-int. by
replacing y with 0
*Inverse of mult. by 2 is div. by 2
|
The x-intercept is (-3, 0).
Next we will find the y- intercept.
What value are we going to plug in for x?
If you said x = 0, you are right. |
 |
*Find y-int. by
replacing x with 0
*Inverse of mult. by -3 is div. by -3
|
| The y-intercept is (0, 2) |
| Step 2: Find
at least one more point.
We can plug in any x value we want as long
as we get the right corresponding y value and
the function exists there.
Let's put in an easy number x =
1: |
 |
*Replace x with
1
*Inverse of add 2 is sub. 2
*Inverse of mult. by -3 is div. by -3
|
| So the ordered pair (1, 8/3) is another solution to our function.
Note that we could have plugged in any value for x: 5, 10, -25, ...,
but it is best to keep it as simple as possible.
The solutions that we found are:
|
x
|
y
|
(x, y)
|
|
-3
|
0
|
(-3, 0)
|
|
0
|
2
|
(0, 2)
|
|
1
|
8/3
|
(1, 8/3)
|
|
| Answer/Discussion
to 1b
x = 3y |
 |
*Find x-int. by
replacing y with 0 |
The x-intercept is (0, 0).
Next we will find the y- intercept.
What value are we going to plug in for x?
If you said, x = 0 you are right. |
 |
*Find y-int. by
replacing x with 0
|
| The y-intercept is (0, 0) |
| Step 2: Find
at least one more point.
Since we really have found only one point this time, we better find
two additional solutions so we have a total of three points.
We can plug in any x value we want as long
as we get the right corresponding y value and
the function exists there.
Let's put in an easy number x =
1: |
 |
*Replace x with
1
*Inverse of mult. by 3 is div. by 3
|
So the ordered pair (1, 1/3) is another solution to our function.
Let's put in another easy number x = -1: |
So the ordered pair (-1, -1/3) is another solution to our function.
The solutions that we found are:
|
x
|
y
|
(x, y)
|
|
0
|
0
|
(-3, 0)
|
|
1
|
1/3
|
(1, 1/3)
|
|
-1
|
-1/3
|
(-1, -1/3)
|
|
| Answer/Discussion
to 2a
x = 4 |
This is in the form x = c.
So, what type of line are we going to end up with?
Vertical. |
Step 1: Find
the x- and y- intercepts.
AND
Step 2: Find
at least one more point.
Since this is a special type of line, I thought I would talk about steps
1 and 2 together.
It does not matter what y is, as long as
x
is 4.
Note that the x-intercept is at (4, 0).
Do we have a y-intercept?
The answer is no. Since x can never
equal 0, then there will be no y-intercept for this equation.
Some points that would be solutions are (4, 0), (4, 1), and (4, 2).
Again, I could have picked an infinite number of solutions.
The solutions that we found are:
|
x
|
y
|
(x, y)
|
|
4
|
0
|
(4, 0)
|
|
4
|
1
|
(4, 1)
|
|
4
|
2
|
(4, 2 )
|
|
| Answer/Discussion
to 2b
y + 5 = 0 |
If you subtract 5 from both sides, you will have y
= -5. It looks like it fits the form y
= c.
With that in mind, what kind of line are we going to end up with?
Horizontal. |
Step 1: Find
the x- and y- intercepts.
AND
Step 2: Find
at least one more point.
Since this is a special type of line, I thought I would talk about steps
1 and 2 together.
It doesn't matter what x is, y
is always -5. So for our solutions we just need three ordered pairs
such that y = -5.
Note that the y-intercept (where x
=
0) is at (0, -5).
Do we have a x-intercept? The answer is no. Since y
has to be -5, then it can never equal 0, which is the criteria of an x-intercept.
So some points that we can use are (0, -5), (1, -5) and (2, -5).
These are all ordered pairs that fit the criteria of y
having to be -5.
Of course, we could have used other solutions, there are an infinite
number of them.
The solutions that we found are:
|
x
|
y
|
(x, y)
|
|
0
|
-5
|
(0, -5)
|
|
1
|
-5
|
(1, -5)
|
|
-1
|
-5
|
(1, -5)
|
|
All contents copyright (C) 2001 - 2008, WTAMU and Kim Seward. All rights reserved.
Last revised on Jan. 9, 2002 by Kim Seward. |
