All Algebra II Resources
Example Questions
Example Question #1 : Graphing Linear Functions
Which equation best matches the graph of the line shown above?
To find an equation of a line, we will always need to know the slope of that line -- and to find the slope, we need at least two points. It looks like we have (0, -3) and (12,0), which we'll call point 1 and point 2, respectively.
Now we need to plug in a point on the line into an equation for a line. We can use either slope-intercept form or point-slope form, but since the answer choices are in point-slope form, let's use that.
Unfortunately, that's not one of the answer choices. That's because we didn't pick the same point to substitute into our equation as the answer choices did. But we can see if any of the answer choices are equivalent to what we found. Our equation is equal to:
which is the slope-intercept form of the line. We have to put all the other answer choices into slope-intercept to see if they match. The only one that works is this one:
Example Question #1 : Graphing Linear Functions
Determine where the graphs of the following equations will intersect.
We can solve the system of equations using the substitution method.
Solve for in the second equation.
Substitute this value of into the first equation.
Now we can solve for .
Solve for using the first equation with this new value of .
The solution is the ordered pair .
Example Question #1 : Graphing Linear Functions
Refer to the line in the above diagram. It we were to continue to draw it so that it intersects the -axis, where would its -intercept be?
First, we need to find the slope of the line.
In order to move from the lower left point to the upper right point, it is necessary to move up five units and right three units. This is a rise of 5 and a run of 3. makes the slope of the line shown .
We can use this to find the -intercept using the slope formula as follows:
The lower left point has coordinates . Therefore, we can set up and solve for in this slope formula, setting :
Example Question #2 : Graphing Linear Functions
Line includes the points and . Line includes the points and . Which of the following statements is true of these lines?
The lines are parallel.
Insufficient information is given to answer this question.
The lines are perpendicular.
The lines are identical.
The lines are distinct but neither parallel nor perpendicular.
The lines are parallel.
We calculate the slopes of the lines using the slope formula.
The slope of line is
The slope of line is
The lines have the same slope, making them either parallel or identical.
Since the slope of each line is 0, both lines are horizontal, and the equation of each takes the form , where is the -coordinate of each point on the line. Therefore, line and line have equations and .This makes them parallel lines.
Example Question #1 : Graphing Linear Functions
What is the equation of the above line?
The equation of a line is with m being the slope and b being the y intercept. The y-intercept is at , so . The x-intercept is , so after plugging in the equation becomes , simplifying to .
Example Question #3 : Graphing Linear Functions
What is the equation of the line displayed above?
The equation of a line is , with m being the slope of the line, and b being the y-intercept. The y intercept of the line is at , so .
The x-intercept is at , the equation becomes , simplification yields
Example Question #3 : Graphing Linear Functions
Refer to the above diagram. which of the following compound inequality statements has this set of points as its graph?
A horizontal line has equation for some value of ; since the line goes through a point with -coordinate 3, the line is . Also, since the line is solid and the region above this line is shaded in, the corresponding inequality is .
A vertical line has equation for some value of ; since the line goes through a point with -coordinate 4, the line is . Also, since the line is solid and the region right of this line is shaded in, the corresponding inequality is .
Since only the region belonging to both sets is shaded - that is, their intersection is shaded - the statements are connected with "and". The correct choice is .
Example Question #1 : Graphing Linear Functions
Which of the following inequalities is graphed above?
First, we determine the equation of the boundary line. This line includes points and , so the slope can be calculated as follows:
Since we also know the -intercept is , we can substitute in the slope-intercept form to obtain the equation of the boundary line:
The boundary is included, as is indicated by the line being solid, so the equality symbol is replaced by either or . To find out which one, we can test a point in the solution set - for ease, we will choose :
_____
_____
_____
0 is less than 3 so the correct symbol is .
The inequality is .
Example Question #1 : Graphing Inequalities
Which of the following inequalities is graphed above?
First, we determine the equation of the boundary line. This line includes points and , so the slope can be calculated as follows:
Since we also know the -intercept is , we can substitute in the slope-intercept form to obtain equation of the boundary:
The boundary is excluded, as is indicated by the line being dashed, so the equality symbol is replaced by either or . To find out which one, we can test a point in the solution set - we will choose :
_____
_____
_____
_____
1 is greater than 0 so the correct symbol is
The inequality is
Example Question #9 : Graphing Linear Functions
Which of the following is the function graphed below?
This function is linear (a line), so we must remember that we can represent lines algebraically using y=mx+b, where m is the slope and b is the y-intercept.
Looking at the graph, we can tell immediately that the y-intercept is -5, because the line crosses(intercepts) the y-axis at -5.
To find the slope, we need two points, and the following formula:
.
For the sake of the example, choose (0,-5) and (2,-1). We can see that the graph clearly passes through each of these points. Any two points will do, however. Substituting each of the values into the slope formula yields m=2.
Thus, our final answer is