In Section 1 we reviewed several ways of making graphs of both lines (by hand) and general functions (using technology).
Example13.8.1.Graphing Lines by Plotting Points.
Graph the equation by creating a table of values and plotting those points.
Explanation.
To make a good table for this line, we should have -values that are multiples of to make sure that the fraction cancels nicely for the outputs.
Point
Figure13.8.2.A table of values for
Figure13.8.3.A graph of
Example13.8.4.Graphing Lines in Slope-Intercept Form.
Find the slope and vertical intercept of , and then use slope triangles to find the next two points on the line. Draw the line.
Explanation.
The slope of is , and the vertical intercept is . Starting at , we go down units and right units to reach more points.
From the graph, we can read that two more points that passes through are and .
Figure13.8.5.A graph of
Example13.8.6.Graphing Lines in Point-Slope Form.
From the equation, find the slope and a point on the graph of , and then use slope triangles to find the next two points on the line. Draw the line.
Explanation.
The slope of is and the point on the graph given in the equation is . So to graph , start at , and the go up units and right units (or down left ) to reach more points.
From the graph, we can read that two more points that passes through are and .
Figure13.8.7.A graph of
Example13.8.8.Graphing Lines Using Intercepts.
Use the intercepts of to graph the equation.
Explanation.
To find the -intercept, set and solve for .
The -intercept is the point .
To find the -intercept, set and solve for .
The -intercept is the point .
Next, we just plot these points and draw the line that runs through them.
Figure13.8.9.A graph of
Example13.8.10.Graphing Functions by Plotting Points.
is how cold it feels outside due to the wind. Imagine a chilly 32 °F day with a breeze blowing over the snowy ground. The wind chill, , at this temperature can be approximated by the formula where is the speed of the wind in miles per hour. This formula only approximates the wind chill for reasonable wind values of about 5 mph to 60 mph. Create a table of values rounded to the nearest tenth for the wind chill at realistic wind speeds and make a graph of .
Explanation.
Typical wind speeds vary between and 20 mph, with gusty conditions up to 60 mph, depending on location. A good way to enter the sixth root into a calculator is to recall that .
Point
Interpretation
A 5 mph wind causes a wind chill of 27.0 °F.
A 10 mph wind causes a wind chill of 23.7 °F.
A 20 mph wind causes a wind chill of 19.9 °F.
A 30 mph wind causes a wind chill of 17.5 °F.
A 40 mph wind causes a wind chill of 15.7 °F.
A 50 mph wind causes a wind chill of 14.2 °F.
A 60 mph wind causes a wind chill of 12.9 °F.
Figure13.8.11.A table of values for
With the values in Table 11, we can sketch the graph.
Figure13.8.12.A graph of
Subsection13.8.2Quadratic Graphs and Vertex Form
In Section 2 we covered the use of technology in analyzing quadratic functions, the vertex form of a quadratic function and how it affects horizontal and vertical shifts of the graph of a parabola, and the factored form of a quadratic function.
Example13.8.13.Exploring Quadratic Functions with Graphing Technology.
Use technology to graph and make a table of the quadratic function defined by and find each of the key points or features.
Find the vertex.
Find the vertical intercept.
Find the horizontal intercept(s).
Find .
Solve using the graph.
Solve using the graph.
State the domain and range of the function.
Explanation.
The specifics of how to use any one particular technology tool vary. Whether you use an app, a physical calculator, or something else, a table and graph should look like:
Additional features of your technology tool can enhance the graph to help answer these questions. You may be able to make the graph appear like:
The vertex is .
The vertical intercept is .
The horizontal intercepts are and .
.
The solutions to are the -values where . We graph the horizontal line and find the -values where the graphs intersect. The solution set is .
The solutions are all -values where the function below (or touching) the line . The solution set is .
The domain is and the range is .
Example13.8.14.The Vertex Form of a Parabola.
Recall that the vertex form of a quadratic function tells us the location of the vertex of a parabola.
State the vertex of the quadratic function .
State the vertex of the quadratic function .
Write the formula for a parabola with vertex and .
Write the formula for a parabola with vertex and .
Explanation.
The vertex of the quadratic function is .
The vertex of the quadratic function is .
The formula for a parabola with vertex and is .
The formula for a parabola with vertex and is .
Example13.8.15.Horizontal and Vertical Shifts.
Identify the horizontal and vertical shifts compared with .
.
.
Explanation.
The graph of the quadratic function is the same as the graph of shifted to the left unit and up units.
The graph of the quadratic function is the same as the graph of shifted to the right units and down units.
Example13.8.16.The Factored Form of a Parabola.
Recall that the factored form of a quadratic function tells us the horizontal intercepts very quickly.
Example13.8.17.Solving Quadratic Equations by Completing the Square.
Solve the equations by completing the square.
Explanation.
To complete the square in the equation , we first we will first move the constant term to the right side of the equation. Then we will use Fact 13.3.3 to find to add to both sides.
In our case, , so
or or or
The solution set is .
To complete the square in the equation , we first divide both sides by since the leading coefficient is 4.
Next, we will complete the square. Since , first,
(13.8.1)
and next, squaring that, we have
.(13.8.2)
So we will add from Equation (13.8.2) to both sides of the equation:
Here, remember that we always factor with the number found in the first step of completing the square, Equation (13.8.1).
or or or
The solution set is .
Example13.8.18.Putting Quadratic Functions in Vertex Form.
Write a formula in vertex form for the function defined by .
Explanation.
Before we can complete the square, we will factor the out of the first two terms. Don’t be tempted to factor the out of the constant term.
Now we will complete the square inside the parentheses by adding and subtracting .
Notice that the constant that we subtracted is inside the parentheses, but it will not be part of our perfect square trinomial. In order to bring it outside, we need to multiply it by . We are distributing the to that term so we can combine it with the outside term.
Note that The vertex is .
Example13.8.19.Graphing Quadratic Functions by Hand.
Graph the function defined by by determining its key features algebraically.
Explanation.
To start, we’ll note that this function opens downward because the leading coefficient, , is negative.
Now we will complete the square to find the vertex. We will factor the out of the first two terms, and then add and subtract on the right side.
The vertex is so the axis of symmetry is the line .
To find the -intercept, we’ll replace with or read the value of from the function in standard form:
The -intercept is and we will find its symmetric point on the graph, which is .
Next, we’ll find the horizontal intercepts. We see this function factors so we will write the factored form to get the horizontal intercepts.
The -intercepts are and .
Now we will plot all of the key points and draw the parabola.
Since the equation is a quadratic equation, we again have several options to consider. We will try factoring on this one first after converting it to standard form.
Here, and two numbers that multiply to be but add to be are and .
oror
The solution set is
Since the equation is a rational we first need to cancel the denominators after factoring and finding the least common denominator.
At this point, we note that the least common denominator is . We need to multiply every term by this least common denominator.
We always check solutions to rational equations to ensure we don’t have any “extraneous solutions”.
???
So, the solution set is .
Since the equation is an absolute value equation, we will first isolate the absolute value and then use Equations with an Absolute Value Expression to solve the remaining equation.
ororor
The solution set is .
Since the equation is a radical equation, we will have to isolate the radical (which is already done), then square both sides to cancel the square root. After that, we will solve whatever remains.
We now have a quadratic equation. We will solve by factoring.
oror
Every potential solution to a radical equation should be verified to check for any “extraneous solutions”.
?or??or?noor
So the solution set is
Since the equation is a quadratic equation, we again have several options. Since the variable only appears once in this equation we will use the the square root method to solve.
ororororor
The solution set is .
Since the equation is a radical equation, we will isolate the radical (which is already done) and then raise both sides to the third power to cancel the cube root.
The solution set is .
Since
is a system of linear equations, we can either use substitution or elimination to solve. Here we will use substitution. To use substitution, we need to solve one of the equations for one of the variables. We will solve the top equation for .
Now, we substitute where ever we see in the other equation.
Now that we have found , we can substitute that back into one of the equations to find . We will substitute into the first equation.
So, the solution must be the point .
Since the equation is quadratic and we are instructed to solve by using completing the square, we should recall that Fact 13.3.3 tells us how to complete the square, after we have sufficiently simplified. Since our equation is already in a simplified state, we need to add to both sides of the equation.
Draw a representation of the union of the sets and .
Explanation.
First we make a number line with both intervals drawn to understand what both sets mean.
Figure13.8.26.A number line sketch of as well as
The two intervals should be viewed as a single object when stating the union, so here is the picture of the union. It looks the same, but now it is a graph of a single set.
Figure13.8.27.A number line sketch of
Example13.8.28.“Or” Compound Inequalities.
Solve the compound inequality.
or
Explanation.
First we will solve each inequality for .
or or or
The solution set to the compound inequality is:
Example13.8.29.Three-Part Inequalities.
Solve the three-part inequality .
Explanation.
This is a three-part inequality. The goal is to isolate in the middle and whatever you do to one “side,” you have to do to the other two “sides.”
The solutions to the three-part inequality are those numbers that are trapped between and , including but not . The solution set in interval notation is .
Example13.8.30.Solving “And” Inequalities.
Solve the compound inequality.
and
Explanation.
This is an “and” inequality. We will solve each part of the inequality and then combine the two solutions sets with an intersection.
andandandand
The solution set to is and the solution set to is . Shown is a graph of these solution sets.
Figure13.8.31.A number line sketch of and also
Recall that an “and” problem finds the intersection of the solution sets. Intersection finds the -values where the two lines overlap, so the solution to the compound inequality must be
.
Example13.8.32.Application of Compound Inequalities.
Mishel wanted to buy some mulch for their spring garden. Each cubic yard of mulch cost and delivery for any size load was . If they wanted to spend between and , set up and solve a compound inequality to solve for the number of cubic yards, , that they could buy.
Explanation.
Since the mulch costs per cubic yard and delivery is , the formula for the cost of yards of mulch is . Since Mishel wants to spend between and , we just trap their cost between these two values.
Note: these values are approximate
Most companies will only sell whole number cubic yards of mulch, so we have to round appropriately. Since Mishel wants to spend more than , we have to round our lower value from up to cubic yards.
If we round the up to , then the total cost will be (which represents ), which is more than Mishel wanted to spend. So we actually have to round down to cubic yards to stay below the maximum.
In conclusion, Mishel could buy ,,, or cubic yards of mulch to stay between and .
Subsection13.8.7Solving Inequalities Graphically
Example13.8.33.Solving Absolute Value Inequalities Graphically.
Solve the inequality graphically.
Explanation.
To solve the inequality , we will start by making a graph with both and .
Figure13.8.34. and
The portion of the graph of that is below is highlighted and the -values of that highlighted region are trapped between and :. That means that the solution set is . Note that we shouldn’t include the endpoints of the interval because at those values, the two graphs are equal whereas the original inequality was only less than and not equal.
Figure 36 shows a graph of . Use the graph do the following.
Solve .
Solve .
Solve .
Figure13.8.36.Graph of
Explanation.
To solve , we first draw a dotted line (since it’s a less-than, not a less-than-or-equal) at . Then we examine the graph to find out where the graph of is underneath the line . Our graph is below the line for -values less than . So the solution set is .
Figure13.8.37.Graph of and solution set to
To solve , we first draw a solid line (since it’s a greater-than-or-equal) at . Then we examine the graph to find out what parts of the graph of are above the line . Our graph is above (or on) the line for -values between and as well as -values bigger than . So the solution set is .
Figure13.8.38.Graph of and solution set to
To solve , we first draw a solid line at and dotted line at . Then we examine the graph to find out what parts of the graph of are trapped between the two lines we just drew. Our graph is between those values for -values between and as well as -values between and as well as as well as -values between and . We use the solid and hollow dots on the graph to decide whether or not to include those values. So the solution set is .
Figure13.8.39.Graph of and solution set to
Exercises13.8.8Exercises
Overview of Graphing.
1.
Create a table of ordered pairs and then make a plot of the equation .
2.
Create a table of ordered pairs and then make a plot of the equation .
3.
Graph the equation .
4.
Graph the equation .
5.
Graph the linear equation by identifying the slope and one point on this line.
6.
Graph the linear equation by identifying the slope and one point on this line.
7.
Make a graph of the line .
8.
Make a graph of the line .
9.
Create a table of ordered pairs and then make a plot of the equation .
10.
Create a table of ordered pairs and then make a plot of the equation .
Quadratic Graphs and Vertex Form.
11.
Let . Use technology to find the following.
The vertex is .
The -intercept is .
The -intercept(s) is/are .
The domain of is .
The range of is .
Calculate . .
Solve .
Solve .
12.
Let . Use technology to find the following.
The vertex is .
The -intercept is .
The -intercept(s) is/are .
The domain of is .
The range of is .
Calculate . .
Solve .
Solve .
13.
An object was launched from the top of a hill with an upward vertical velocity of feet per second. The height of the object can be modeled by the function , where represents the number of seconds after the launch. Assume the object landed on the ground at sea level. Find the answer using technology.
seconds after its launch, the object reached its maximum height of feet.
14.
An object was launched from the top of a hill with an upward vertical velocity of feet per second. The height of the object can be modeled by the function , where represents the number of seconds after the launch. Assume the object landed on the ground at sea level. Find the answer using technology.
seconds after its launch, the object fell to the ground at sea level.
15.
Consider the graph of the equation . Compared to the graph of , the vertex has been shifted units
?
left
right
and units
?
down
up
.
16.
Consider the graph of the equation . Compared to the graph of , the vertex has been shifted units
?
left
right
and units
?
down
up
.
17.
The quadratic expression is written in vertex form.
Write the expression in standard form.
Write the expression in factored form.
18.
The quadratic expression is written in vertex form.
Write the expression in standard form.
Write the expression in factored form.
19.
The formula for a quadratic function is .
The -intercept is .
The -intercept(s) is/are .
20.
The formula for a quadratic function is .
The -intercept is .
The -intercept(s) is/are .
Completing the Square.
21.
Solve the equation by completing the square.
22.
Solve the equation by completing the square.
23.
Solve the equation by completing the square.
24.
Solve the equation by completing the square.
25.
Complete the square to convert the quadratic function from standard form to vertex form, and use the result to find the function’s domain and range.
The domain of is
The range of is
26.
Complete the square to convert the quadratic function from standard form to vertex form, and use the result to find the function’s domain and range.
The domain of is
The range of is
27.
Graph by algebraically determining its key features. Then state the domain and range of the function.
28.
Graph by algebraically determining its key features. Then state the domain and range of the function.
29.
Graph by algebraically determining its key features. Then state the domain and range of the function.
30.
Graph by algebraically determining its key features. Then state the domain and range of the function.
Absolute Value Equations.
31.
Solve the equation.
32.
Solve the equation.
33.
Solve the equation.
34.
Solve the equation.
35.
Solve the equation.
36.
Solve the equation.
37.
Solve the equation.
38.
Solve the equation.
Exercise Group.
39.
Solve the equation.
40.
Solve the equation.
41.
Solve the equation.
42.
Solve the equation.
Solving Mixed Equations.
43.
Solve the equation.
44.
Solve the equation.
45.
Solve the equation.
46.
Solve the equation.
47.
Solve the equation.
48.
Solve the equation.
49.
Solve the equation.
50.
Solve the equation.
51.
Solve the equation.
52.
Solve the equation.
53.
Solve the equation by completing the square.
54.
Solve the equation by completing the square.
Compound Inequalities.
55.
Solve the compound inequality algebraically.
or
56.
Solve the compound inequality algebraically.
or
57.
Solve the compound inequality algebraically.
and
58.
Solve the compound inequality algebraically.
and
59.
Solve the compound inequality algebraically.
and
60.
Solve the compound inequality algebraically.
and
61.
Solve the compound inequality algebraically.
and
62.
Solve the compound inequality algebraically.
or
63.
Solve the compound inequality algebraically.
64.
Solve the compound inequality algebraically.
65.
Solve the compound inequality algebraically.
is in
66.
Solve the compound inequality algebraically.
is in
Solving Inequalities Graphically.
67.
Solve the equations and inequalities graphically. Use interval notation when applicable.
68.
Solve the equations and inequalities graphically. Use interval notation when applicable.
Exercise Group.
69.
The equations and are plotted.
What are the points of intersection?
Solve .
Solve .
70.
The equations and are plotted.
What are the points of intersection?
Solve .
Solve .
71.
The equations and are plotted.
What are the points of intersection?
Solve .
Solve .
72.
The equations and are plotted.
What are the points of intersection?
Solve .
Solve .
73.
The equations and are plotted.
What are the points of intersection?
Solve .
Solve .
74.
Use graphing technology to solve the inequality . State the solution set using interval notation, and approximate if necessary.
Exercise Group.
75.
A graph of is given. Use the graph alone to solve the compound inequalities.
76.
A graph of is given. Use the graph alone to solve the compound inequalities.
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