College Algebra Tutorial 27

College Algebra
Answer/Discussion to Practice Problems
Tutorial 27: Graphing Lines

WTAMU > Virtual Math Lab > College Algebra > Tutorial 27: Graphing Lines

Answer/Discussion to 1a

Step 1: Put the equation in slope/intercept form () and identify the slope and y-intercept.

Solve for y to get it into the slope/intercept form:

*Inverse of add. 10x is sub. 10x

*Inverse of mult. -5 is div. -5

*Slope/intercept form

Lining up the form with the equation we got, can you see what the slope and y-intercept are?

In this form, the slope is m, which is the number in front of x. In our problem that would have to be 2.

In this form, the y-intercept is b, which is the constant. In our problem that would be -2.

How did you do?

Step 2: Plot the y-intercept on a 2-dimensional graph.

Since the y-intercept is where the line crosses the y-axis, then x's value would have to be 0. In step 1 we found our y-intercept value to be -2.

Putting that together, the ordered pair for the y-intercept would be (0, -2):
ad1a2

Step 3: Use the slope to find a second point on the line.

In step 1 we found our slope to be 2.

Since we have a positive slope the rise and the run need to either be BOTH positive or BOTH negative. So, we can either rise up 2 and run right 1 OR go down 2 and left 2.

I chose to rise up 2 and run right 1, starting on the y-intercept:
ad1a3

Note that if we would have gone down 2 and left 1 from our y-intercept, that we would have ended up at (-1, -4) which would have lined up with the other points.

Step 4: Draw a line through the points found in steps 2 and 3.

(return to problem 1a)

Answer/Discussion to 1b

Step 1: Put the equation in slope/intercept form () and identify the slope and y-intercept.

This equation is already in slope/intercept form:

*Slope/intercept form

Lining up the form with the equation we got, can you see what the slope and y-intercept are?

In this form, the slope is m, which is the number in front of x. In our problem that would have to be -2/3.

In this form, the y-intercept is b, which is the constant. In our problem that would be 0.

How did you do?

Step 2: Plot the y-intercept on a 2-dimensional graph.

Since the y-intercept is where the line crosses the y-axis, then x's value would have to be 0. In step 1 we found our y-intercept value to be 0.

Putting that together, the ordered pair for the y-intercept would be (0, 0):
ad1b2

Step 3: Use the slope to find a second point on the line.

In step 1 we found our slope to be -2/3.

Since we have a negative slope the rise and the run have to be opposites of each other, one has to be positive and one has to be negative. So, we can either go down 2 and run right 3 OR rise up 2 and run left 3.

I chose to go down 2 and run right 3, starting on the y-intercept:
ad1b3

Note that if we would have rose up 2 and ran left 3 from our y-intercept, that we would have ended up at (-3, 2) which would have lined up with the other points.

Step 4: Draw a line through the points found in steps 2 and 3.

(return to problem 1b)

Answer/Discussion to 1c

Note how we have an equation that has x set equal to a constant, where there is no y. Since this fits the form of x = c described above, we can jump to the chase and draw our vertical line:

Since we have a vertical line, what is our slope going to be? If you said undefined, you are so right!!!

What would the y-intercept be? Give yourself a high five if you said there is no y-intercept.

(return to problem 1c)

Answer/Discussion to 1d

Note how we are missing an x and even though it is not quite in the form y = c, that we can get it in that form.

Lets first rewrite this in the form y = c and then go from there:

*Written in the form y = c

Note how we have an equation that has y set equal to a constant, where there is no x. Since this fits the form of y = c described above, we can jump to the chase and draw our horizontal line:

Since we have a horizontal line, what is our slope going to be? If you said 0, you are so right!!!

What would the y-intercept be? Give yourself a high five if you said -4.

(return to problem 1d)

WTAMU > Virtual Math Lab > College Algebra > Tutorial 27: Graphing Lines