# Algebra II : Graphing Linear Functions

## Example Questions

### Example Question #1 : Graphing Linear Functions

Select the equation of the line perpendicular to the graph of .

None of these.

Explanation:

Lines are perpendicular when their slopes are the negative recicprocals of each other such as . To find the slope of our equation we must change it to slope y-intercept form.

Subtract the x variable from both sides:

Divide by 4 to isolate y:

The negative reciprocal of the above slope:  . The only equation with this slope is

### Example Question #11 : Graphing Linear Functions

Where does  cross the  axis?

Explanation:

To find where this equation crosses the  axis or its -intercept, change the equation into slope intercept form.

Subtract to isolate :

Divide both sides by  to completely isolate :

This form is the slope intercept form  where  is the slope of the line and  is the -intercept.

### Example Question #12 : Graphing Linear Functions

Find the -intercepts and the -intercepts of the equation.

and

and

and

and

and

Explanation:

To find the x-intercepts, remember that the line is crossing the x-axis, and that y=0 when the line crosses the x-axis.

So plug in y=0 into the equation above.

To find the y-intercepts, remember that the line is crossing the y-axis, and that x=0 when the line crosses the x-axis.

So plug in x=0 into the equation above.

### Example Question #13 : Graphing Linear Functions

Find the slope of the line that passes through the pair of points. Express the fraction in simplest form.

and

Explanation:

Slope is the change of a line. To find this line one can remember it as rise over run. This rise over run is really the change in the y direction over the change in the x direction.

Therefore the formula for slope is as follows.

Plugging in our given points

and

### Example Question #14 : Graphing Linear Functions

Which of the following equations passes through  and is parallel to .

Explanation:

Since the line goes through  we know that  is the y-intercept.

Since we are looking for parallel lines, we need to write the equation of a line that has the same slope as the original, which is .

Slope-intercept form equation is , where  is the slope and  is the y-intercept.

Therefore,

.

### Example Question #15 : Graphing Linear Functions

Write an equation of the line passing through  and  in slope-intercept form.

Reminder: Slope-Intercept form is , where  is the slope and  is the y-intercept.

Explanation:

Step 1: Find the Slope

Step 2: Find the y-intercept

Use the slope and a point in the original y-intercept

### Example Question #16 : Graphing Linear Functions

Find the slope-intercept form of an equation of the line that has a slope of  and passes through .

Explanation:

Since we know the slope and we know a point on the line we can use those two piece of information to find the y-intercept.

### Example Question #17 : Graphing Linear Functions

Determine the slope of a line that has points  and .

Explanation:

Slope is the change of a line. To find this line one can remember it as rise over run. This rise over run is really the change in the y direction over the change in the x direction.

Therefore the formula for slope is as follows.

Plugging in our given points

and ,

.

### Example Question #18 : Graphing Linear Functions

What is the equation of the line passing through (-1,4) and (2,6)?

Explanation:

To find the equation of this line, first find the slope. Recall that slope is the change in y over the change in x: . Then, pick a point and use the slope to plug into the point-slope formula (): . Distribute and simplify so that you solve for y: .

### Example Question #2 : Graphing Linear Functions

An individual's maximum heart rate can be found by subtracting his or her age from . Which graph correctly expresses this relationship between years of age and maximum heart rate?