All Algebra II Resources
Example Questions
Example Question #11 : Graphing Linear Functions
Select the equation of the line perpendicular to the graph of .
None of these.
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?
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 #851 : Algebra Ii
Find the -intercepts and the -intercepts of the equation.
and
and
and
and
and
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 #34 : Linear Functions
Find the slope of the line that passes through the pair of points. Express the fraction in simplest form.
and
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 #11 : Graphing Linear Functions
Which of the following equations passes through and is parallel to .
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 #42 : 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.
Step 1: Find the Slope
Step 2: Find the y-intercept
Use the slope and a point in the original y-intercept
Step 3: Write your equation
Example Question #41 : Linear Functions
Find the slope-intercept form of an equation of the line that has a slope of and passes through .
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 #12 : Graphing Linear Functions
Determine the slope of a line that has points and .
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)?
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?
In form, where y = maximum heart rate and x = age, we can express the relationship as:
We are looking for a graph with a slope of -1 and a y-intercept of 220.
The slope is -1 because as you grow one year older, your maximum heart rate decreases by 1.
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