Topics
3. Linear Functions and Equations
topic badge
United States of America
Grades 9-12

3.06 Forms of linear functions

Lesson
Practice

Introduction

There are three main forms of a linear function: slope-intercept form, standard form, and point-slope form. Each form is based on a different set of parameters and all forms can be used to identify and interpret the key features of a linear function.

Slope-intercept form

Exploration

In the applet, move the m slider and the b slider.

Loading interactive...
  1. What happens to the graph when the value of m changes?
  2. What happens to the graph when the value of b changes?

The slope, m, changes the steepness of the line.

The y-intercept, b, changes where the line crosses the y-axis.

This allows us to write the equation of a line in slope-intercept form:

\displaystyle y=mx+b
\bm{m}
slope
\bm{b}
y-intercept

The main advantage of slope-intercept form is that we can easily identify two key features: the slope and the y-intercept directly from the equation. This form is especially helpful when we want to graph a linear function.

With equations in slope-intercept form, we can efficiently identify two points that define the line and allow us to sketch the graph:

  1. Plot the y-intercept (0,b).
  2. Write the slope m as a fraction, where the numerator will be the change in y-values or the rise, and the denominator will be the change in x-values or the run.
  3. From the y-intercept as the initial value, count the rise and run on the graph to the next point on the graph and connect the points, drawing a line through the graph.

Examples

Example 1

Consider the following equation:

2y=-x+12

a

Write the equation in slope-intercept form.

Worked Solution
Create a strategy

Since the slope-intercept form of a linear equation is the form y=mx+b, we should apply properties of equality to manipulate the equation to this form.

Apply the idea
\displaystyle 2y\displaystyle =\displaystyle -x+12Original equation
\displaystyle \dfrac{2y}{2}\displaystyle =\displaystyle \dfrac{-x}{2}+\dfrac{12}{2}Divide by 2
\displaystyle y\displaystyle =\displaystyle -\dfrac{1}{2}x+6Evaluate the division
Reflect and check

With the equation now in slope-intercept form, we can see that the slope of the function is -\dfrac{1}{2} and the y-intercept of the function is 6.

b

Graph the equation.

Worked Solution
Create a strategy

Once the equation is in slope-intercept form, we can plot the y-intercept and follow the slope to the next point on the graph.

Apply the idea

Based on the equation y=-\dfrac{1}{2}x+6, the y-intercept is (0,6).

-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
x
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
y

The slope from the equation is -\dfrac{1}{2}, so we can label it \dfrac{-1}{2}=\dfrac{\text{rise}}{\text{run}}. Since the change in y-values or rise is -1, we will move down the graph from (0,6) one space. Then, since the change in x-values or run is 2, we will move to the right 2 spaces and plot a point.

-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
x
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
y

Finally, we can use the two points we plotted and draw a line through the graph.

-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
x
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
y
Reflect and check

After plotting the first point on the graph, we can continue counting the rise and run to follow the pattern through the graph. The line will be more accurate by plotting more points before drawing the line.

-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
x
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
y
c

Is the point (8, 2) a solution to the equation?

Worked Solution
Create a strategy

We can substitute the ordered pair into the equation of the function and determine whether the statement is true. Then, we can confirm whether the point is a solution to the equation and lies on the line formed by the equation.

Apply the idea

We have

\displaystyle 2y\displaystyle =\displaystyle -x+12Original equation
\displaystyle 2(2)\displaystyle =\displaystyle -(8)+12Substitute y=2 and x=8
\displaystyle 4\displaystyle =\displaystyle 4Evaluate the multiplication and addition

Since we are left with both sides of the equation being equal, we can say the ordered pair (8, 2) is a solution to the equation.

Reflect and check

We can also extend the graph to see that the point (8,2) is on the line.

-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
7
8
9
x
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
y

Example 2

A bathtub has a clogged drain, so it needs to be pumped out. It currently contains 30 gallons of water.

The table of values shows the linear relationship of the amount of water remaining in the tub, y, after x minutes.

\text{Time in minutes } (x)0123
\text{Water remaining in gallons } (y)30282624
a

Determine the linear equation in slope-intercept form that represents this situation.

Worked Solution
Create a strategy

We can pick two points to calculate the rate of change for the slope. Then we can recognize that the y-intercept is given in the table of values.

Apply the idea

Find the slope using the values \left(0,30\right) and \left(1,28\right):

\displaystyle m\displaystyle =\displaystyle \dfrac{y_2-y_1}{x_2-x_1}Slope formula
\displaystyle m\displaystyle =\displaystyle \dfrac{28-30}{1-0}Substitute \left(x_1,y_1\right)=\left(1,30\right) and \left(x_2,y_2\right)=\left(1,28\right)
\displaystyle m\displaystyle =\displaystyle -2Evaluate the subtraction and division

Notice that the initial value, or y-intercept is given in the table as \left(0,30\right).

Writing in the form y=mx+b the equation that represents this situation is y=-2x+30.

Reflect and check

If we had not noticed that the y-intercept was given, we could have substituted in any pair of values for x and y, and solved for b.

b

Draw the graph of this linear relationship with a clearly labeled scale. Only show the viable solutions.

Worked Solution
Create a strategy

We can't have a negative time (x \geq 0) and we should end the graph when the tub is empty (y=0). To plan our graph, we need to find when the tub is empty.

Apply the idea

To find when the tub is empty:

\displaystyle y\displaystyle =\displaystyle -2x+30Original equation
\displaystyle 0\displaystyle =\displaystyle -2x+30Substitute y=0
\displaystyle -30\displaystyle =\displaystyle -2xSubtraction property of equality
\displaystyle 15\displaystyle =\displaystyle xDivision property of equality

This tells us that we have the restriction that 0 \leq x\leq 15 and 0 \leq y\leq 30, which will help us choose the appropriate axes and scale.

5
10
15
\text{Time in minutes}
5
10
15
20
25
30
\text{Water in gallons}

We know that the bathtub begins with 30 gallons which is represented by the y-intercept at (0,30) and that the slope of the linear equation is -2 which means that the tub loses 2 gallons of water every minute until there is no water in the tub at 15 minutes.

5
10
15
\text{Time in minutes}
5
10
15
20
25
30
\text{Water in gallons}
Reflect and check

We can also graph the slope by using the idea of \dfrac{\text{rise}}{\text{run}}. Since the slope is -2, we can write it as a fraction \dfrac{-2}{1} and identify that the change in y-values (or rise) is -2 and that the change in x-values (or run) is 1.

c

Describe how the graph would change if, instead, there were initially 40 gallons of water in the tub, and it emptied at 2.5 gallons per minute.

Worked Solution
Create a strategy

The initial value is represented by the y-intercept on the graph. The rate of change is represented by the slope of the graph. We should consider how these are changing from the original question.

Apply the idea

The original question had an initial value of 30 gallons, and the new scenario has an initial value of 40 gallons. This means the new y-intercept will be higher on the y-axis.

The original question had a rate of change of -2 as it was decreasing at 2 gallons per minute. The new scenario has a rate of change of -2.5 as it is emptying at 2.5 gallons per minute. This means the slope will be steeper in the new scenario.

We can see this on the graph:

5
10
15
20
\text{Time }(x)
5
10
15
20
25
30
35
40
\text{Water }(y)
Reflect and check

Notice that despite starting with 10 extra gallons of water, the tub with 40 gallons of water only takes 1 more minute to empty than the 30-gallon tub, because it is emptying at a faster rate. This is reflected in the graph as a steeper slope. The second function is decreasing at a greater rate than the first.

Example 3

Imogen is a cyclist. She typically bikes at 15 \text{ mph}. She is doing a 50-mile bike ride for charity.

a

Draw a graph that shows her distance remaining throughout the 50-mile ride if she bikes her typical speed.

Worked Solution
Create a strategy

We need to decide what our axes should represent before we can graph anything.

Since we are given her speed in miles per hour and speed typically is the slope of our graph we know that \dfrac{\text{miles}}{\text{hour}} \to \dfrac{\text{rise}}{\text{run}}, so we can use the units of "miles" for our y-axis and "hours" for our x-axis.

We also need to consider what scale would be appropriate. We know the distance goes from 50 down to 0, so we can use the speed to determine and an appropriate scale for the x-axis.

Apply the idea

We can use a table of values to see when the distance gets down below 0 as we don't need to go beyond that.

Time (hours)01234
Distance remaining (miles)5035205-10
1
2
3
4
\text{Time (hours)}
5
10
15
20
25
30
35
40
45
50
55
\text{Distance remaining (miles)}

This would be an appropriate scale and axes.

1
2
3
4
\text{Time (hours)}
5
10
15
20
25
30
35
40
45
50
55
\text{Distance remaining (miles)}

Initially, when \text{Time}=0, the distance remaining would be 50. This means we have a y-intercept of (0,50). Then using our labeled axes, we can use the rate of change, -15 \text{ mph}, as the slope of the line to graph or use our table of values.

Reflect and check

Notice that we were asked to show the distance remaining in the graph. As Imogen pedals, she gets closer to her destination, so the distance remaining will decrease. This means the rate of change, or slope, must be negative. If we were showing the distance traveled, the distance would be increasing over time and we would have used a positive rate of change.

b

Write the linear equation that represents the graph in part (a).

Worked Solution
Create a strategy

We can use x and y as variables, or any letters of our choice, as long as we state what they represent.

Since this is a linear function, we can write the equation in slope-intercept form.

Apply the idea

Let t represent the time in hours that Imogen has been biking for.

Let d represent the distance in miles remaining after t hours.

Slope: -15

y-intercept: 50

Equation: d=-15t+50

Reflect and check

We can also write the equation as d=50-15t.

c

Explain whether or not we can predict the distance remaining after 5 hours.

Worked Solution
Create a strategy

We can use the equation from part (b) or the graph from part (a), but either way, we need to consider whether or not the solution we find is viable.

Apply the idea
1
2
3
4
5
\text{Time (hours)}
-20
-10
10
20
30
40
50
\text{Distance remaining (miles)}

Looking at the graph and extending it beyond the x-intercept, we can see that for any x-value greater than 3 \dfrac{1}{3}, the distance remaining would be negative. Since 5>3 \dfrac{1}{3}, we would get a negative distance remaining in our prediction.

We can conclude that using this model, we cannot have a viable prediction for the distance remaining after 5 hours.

Reflect and check

If we had used the equation from part (b), we would get a negative answer which is not a viable distance remaining as she stops biking after 50 miles.

\displaystyle d\displaystyle =\displaystyle -15t+50Equation from part (b)
\displaystyle d\displaystyle =\displaystyle -15(5)+50Substitute t=5
\displaystyle d\displaystyle =\displaystyle -75+50Evaluate the product
\displaystyle d\displaystyle =\displaystyle -25Evaluate the sum
Idea summary

The slope-intercept form of a line is:

\displaystyle y=mx+b
\bm{m}
slope
\bm{b}
y-intercept

Slope-intercept form is useful when we know, or want to know the slope of the line and the y-intercept of the line.

Standard form

The standard form of a linear relationship is a way of writing the equation with all of the variables on one side:

\displaystyle Ax+By=C
\bm{A}
a non-negative integer
\bm{B,C}
integers
\bm{A,B}
are not both 0

To draw the graph from standard form, we can find and plot the x- and y-intercepts or convert to slope-intercept form.

x
y

In order to find the x-intercept, substitute 0 for y in the equation and solve:

Ax+B(0)=C

In order to find the y-intercept, substitute 0 for x in the equation and solve:

A(0)+By=C

A special case of standard form is y=b, when m=0. This will be a horizontal line since the rise will be 0. Every y-value will be b for each value of x. The slope of any horizontal line will be zero.

We can also have vertical lines in the form x=a, where every value of x will be a for each value of y. If we calcualted the slope using any two points, we would get a value of 0 in the denominator. Therefore, we say slope of a vertical line is undefined.

x
y
A horizontal line of the form y=b
x
y
A vertical line of the form x=a

Examples

Example 4

Determine whether the equation \dfrac{1}{2}x+4y=16 is written in standard form.

Worked Solution
Create a strategy

In order for the equation to be written in standard form, we must confirm that the equation is written in the form Ax+By=C, that A is a non-negative integer, B and C are both integers, and A and B are not both zero.

Apply the idea

Since A=\dfrac{1}{2} in the equation, it's not written in standard form because A must be an integer and integers are the set of positive and negative whole numbers.

Reflect and check

To convert the equation to standard form, we can use properties of equality and multiply each term by 2 to change the rational coefficient of x.

\displaystyle \dfrac{1}{2}x + 4y\displaystyle =\displaystyle 16Original equation
\displaystyle 2 \cdot (\dfrac{1}{2}x + 4y)\displaystyle =\displaystyle 2 \cdot 16Multiplication property of equality
\displaystyle x +8y\displaystyle =\displaystyle 32Evaluate the multiplication

A=1, B=8, and C=32 now meet the requirements for standard form.

Example 5

A tour company travels to the Great Smoky Mountains National Park. They use a combination of buses and vans to get tourists to their destination. One bus can take 42 passengers, and one van can take 7 passengers. One day, they have 168 people register for the tour.

a

Write an equation in standard form that could be used to model the number of buses and vans they could use to transport all the people registered.

Worked Solution
Create a strategy

In words, we can start with the idea that: \left(\text{Number of people on buses}\right)+\left(\text{Number of people on vans}\right)=\text{Total number of people} and that: \text{Number of people on buses}=42 \cdot \left(\text{Number of buses}\right) and:\text{Number of people on vans}=7 \cdot \left(\text{Number of vans}\right)

We will then need to define variables and write an equation using them.

Apply the idea

Let b represent the number of buses used and let v represent the number of vans used.

So the \text{Number of people on buses}=42b and \text{Number of people on vans}=7v

which finally gives us the whole equation: 42b+7v=168

Reflect and check

It is important to declare variables and it can be helpful to use variables that relate to the quantities in the scenario to make sure we don't mix them up.

b

Graph the equation with an appropriate scale and labels.

Worked Solution
Create a strategy

In this case, there isn't a clear independent and dependent variable, so we can put b on the horizontal axis and v on the vertical axis.

To determine an appropriate scale, we can first find the values of the intercepts as those can give an idea of the maximum values for each axis.

Apply the idea

Find the b-intercept:

\displaystyle 42b+7v\displaystyle =\displaystyle 168Equation from part (a)
\displaystyle 42b+7(0)\displaystyle =\displaystyle 168Substitute v=0
\displaystyle 42b\displaystyle =\displaystyle 168Evaluate the product
\displaystyle b\displaystyle =\displaystyle 4Division property of equality

Find the v-intercept:

\displaystyle 42b+7v\displaystyle =\displaystyle 168Equation from part (a)
\displaystyle 42(0)+7v\displaystyle =\displaystyle 168Substitute b=0
\displaystyle 7v\displaystyle =\displaystyle 168Evaluate the product
\displaystyle v\displaystyle =\displaystyle 24Division property of equality
1
2
3
4
5
\text{Number of buses }\left(b\right)
2
4
6
8
10
12
14
16
18
20
22
24
26
\text{Number of vans }\left(v\right)

An appropriate scale would be going up by 1 along the b-axis to a maximum of 5 and going up by 2 or 4 along the v-axis to a maximum of 26. This will extend both of our axes just past where we need to plot the intercepts.

Reflect and check

The standard form is helpful when looking at scenarios that have a mixture of two different items.

When we identify the intercepts in a mixture scenario, it can be interpreted as the amount of that item when none of the other items is included.

In this case, if we made b the independent variable instead, the graph would be like this:

4
8
12
16
20
24
\text{Number of vans }\left(v\right)
1
2
3
4
5
\text{Number of buses }\left(b\right)
c

Predict the number of vans that would be required if only 1 bus was available.

Worked Solution
Create a strategy

We can find the point along the b-axis where b=1 and then up to the line and across to the v-axis to find the corresponding value for v, the number of vans.

Apply the idea
1
2
3
4
5
\text{Number of buses }\left(b\right)
2
4
6
8
10
12
14
16
18
20
22
24
26
\text{Number of vans }\left(v\right)

Using the graph, we can see that the point (1,18) lies on the graph of the equation. If only 1 bus was available, they would need 18 vans.

Reflect and check

We can check using the equation.

\displaystyle 42b+7v\displaystyle =\displaystyle 168Equation from part (a)
\displaystyle 42(1)+7v\displaystyle =\displaystyle 168Substitute b=1
\displaystyle 42+7v\displaystyle =\displaystyle 168Evaluate the product
\displaystyle 7v\displaystyle =\displaystyle 126Subtraction property of equality
\displaystyle v\displaystyle =\displaystyle 18Division property of equality
Idea summary

The standard form of a line is:

\displaystyle Ax+By=C
\bm{A}
a non-negative integer
\bm{B,C}
integers
\bm{A,B}
are not both 0

Standard form is useful when we know, or want to know both intercepts of the line.

Point-slope form

When we are given the coordinates of a point on the line and the slope of that line, then the point-slope form can be used to state the equation of the line.

Coordinates can be given as an ordered pair in a table of values, read from a graph, or described in a scenario. The slope can be stated as a value, calculated from two points, read from a graph, or given as a rate of change in a scenario.

Point-slope form of a linear relationship:

\displaystyle y-y_1=m\left(x-x_1\right)
\bm{m}
slope
\bm{x_1}
x-coordinate of a point on the line
\bm{y_1}
y-coordinate of the same point

We can also find the equation in point-slope form when given the coordinates of two points on the line by first finding the slope of the line.

We now have three forms of expressing the same equation, and each provides us with useful information about the line formed.

Exploration

Drag the point on the graph to move the line and drag the slider to change the slope.

Loading interactive...
  1. If we are given a point on the line and the slope of the line, what is the best way to find the equation of the line?
  2. In what ways are the slope formula and the point-slope form connected to one another?

If we are given a point on the line and the slope of the line, we can find the equation of the line by substituting the values of the point and slope into the point-slope form of a linear equation. The slope-intercept form of the equation can be found by solving the point-slope form for y.

Examples

Example 6

For each of the following equations, determine if they are in point-slope form, standard form, or slope-intercept form. If they are not in standard form, convert them to standard form.

a

y-3=-\dfrac{2}{5}\left(x+7\right)

Worked Solution
Apply the idea

The given equation is in point-slope form.

\displaystyle y-3\displaystyle =\displaystyle -\dfrac{2}{5}\left(x+7\right)Original equation
\displaystyle 5y-15\displaystyle =\displaystyle -2\left(x+7\right)Multiplication property of equality
\displaystyle 5y-15\displaystyle =\displaystyle -2x-14Distributive property
\displaystyle 2x+5y\displaystyle =\displaystyle 1Addition property of equality

2x+5y=1 is in standard form.

Reflect and check

There are more steps to convert from point-slope form to standard form than from point-slope form to slope-intercept form.

b

y=4x-10

Worked Solution
Apply the idea

The given equation is in slope-intercept form.

\displaystyle y\displaystyle =\displaystyle 4x-10Original equation
\displaystyle -4x+y\displaystyle =\displaystyle -10Subtraction property of equality
\displaystyle 4x-y\displaystyle =\displaystyle 10Division property of equality

4x-y=10 is in standard form.

Example 7

A line passes through the two points \left(-3,7\right) and \left(2,-3\right).

a

Write the equation of the line in point-slope form.

Worked Solution
Create a strategy

We will first find the slope of the line using the two points. Then we will pick one of the points to substitute into the point-slope equation.

Apply the idea

Find the slope of the line:

\displaystyle m\displaystyle =\displaystyle \dfrac{y_2-y_1}{x_2-x_1}Slope formula
\displaystyle m\displaystyle =\displaystyle \dfrac{-3-7}{2-\left(-3\right)}Substitute \left(x_1,y_1\right)=\left(-3,7\right) and \left(x_2,y_2\right)=\left(2,-3\right)
\displaystyle m\displaystyle =\displaystyle \dfrac{-10}{5}Evaluate the subtraction
\displaystyle m\displaystyle =\displaystyle -2Evaluate the division

Find the equation of the line in point-slope form:

\displaystyle y-y_1\displaystyle =\displaystyle m\left(x-x_1\right)Point-slope formula
\displaystyle y-7\displaystyle =\displaystyle -2\left(x-\left(-3\right)\right)Substitute y_1=7, m=-2, and x_1=-3
\displaystyle y-7\displaystyle =\displaystyle -2\left(x+3\right)Simplify
Reflect and check

We can confirm we have correctly written this in point-slope form: y-y_1=m\left(x-x_1\right), as we can see the coordinates \left(-3,7\right), and a slope of -2 are represented correctly. This is easier to see in the previous line: y-7=-2\left(x-\left(-3\right)\right)

b

Determine whether the ordered pair (-10,21) lies on the same line as \left(-3,7\right) and \left(2,-3\right).

Worked Solution
Create a strategy

We can substitute the ordered pair into the equation we wrote in part (a) and determine whether the statement is true. If so, we can confirm whether the point lies on the same line as \left(-3,7\right) and \left(2,-3\right).

Apply the idea

We have

\displaystyle y-7\displaystyle =\displaystyle -2(x+3)Point-slope form from part (a)
\displaystyle (21)-7\displaystyle =\displaystyle -2((-10)+3)Substitute y=21 and x=-10
\displaystyle 14\displaystyle =\displaystyle -2(-7)Evaluate the subtraction and addition
\displaystyle 14\displaystyle =\displaystyle 14Evaluate the multiplication

Since the resulting equation is true, the ordered pair (-10, 21) satisfies the equation and as a result is on the same line as \left(-3,7\right) and \left(2,-3\right).

Reflect and check

Any ordered pair that satisfies the equation will be on the same line as \left(-3,7\right) and \left(2,-3\right).

Example 8

A carpenter charges for a day's work using the given equation, where y is the cost and x is the number of hours worked:y-125=50\left(x-2\right)

a

Draw the graph of the linear equation from the point-slope form. Clearly label the axes with labels, units, and an appropriate scale.

Worked Solution
Create a strategy

We need two points to plot a line. We can read from the equation that one point on the line is \left(2,125\right). To get another point, we can use the slope from the given point or substitute in an x-value in the domain and solve for y.

Since the x-axis will be the number of hours worked and generally an 8-hour work day is reasonable, showing the graph for 0\leq x\leq 8 would be a good scale.

Apply the idea

Before we graph, we label our axes. To do that, we need to know what our maximum and minimum values should be.

We can use the slope or the equation to determine the y-value when x=8. We can find that the y-value when x=8 is y=425.

1
2
3
4
5
6
7
8
\text{Hours }(x)
50
100
150
200
250
300
350
400
450
\text{Charge in }\$ \,(y)

A reasonable scale for the x-axis from 0 to 8 going up by 1.

A reasonable scale for the y-axis from 0 to 450 going up by 50 with one tick between each label at the 25s.

1
2
3
4
5
6
7
8
\text{Hours } (x)
50
100
150
200
250
300
350
400
450
\text{Charge in }\$ \,(y)

The given point is \left(2,125\right).

Since the slope is 50, we can go up 50 units and right 1 unit from our given point to plot another point on the line.

b

Predict the charge for 6 hours of work using the graph.

Worked Solution
Create a strategy

We can go to x=6 on the x-axis and then go up to the line and across to the y-axis to make our prediction.

Apply the idea
1
2
3
4
5
6
7
8
\text{Hours } (x)
50
100
150
200
250
300
350
400
450
\text{Charge in }\$ \,(y)

The charge for 6 hours of work will be \$ 325.

c

Give an example of a non-viable solution if the carpenter only uses this model for a maximum of 10 hours per day. Explain your answer.

Worked Solution
Create a strategy

Since the carpenter only uses this model between 0 and 10 hours, any solution outside of this interval will not be viable.

Apply the idea

One possible non-viable solution would be \left(11, 575\right) which would represent working 11 hours and getting paid \$575, but the model only applies to a maximum of 10 hours, so we don't actually know what would happen for 11 hours.

Reflect and check

Any solution with x<0 or x>10 would be non-viable.

Idea summary

The point-slope form of a line is:

\displaystyle y-y_1=m\left(x-x_1\right)
\bm{m}
slope
\bm{x_1}
x-coordinate of a point on the line
\bm{y_1}
y-coordinate of the same point

Point-slope form is useful when we know or want to know the slope of the line and a point on the line.

Outcomes

A.SSE.A.2

Use the structure of an expression to identify ways to rewrite it.

A.CED.A.2

Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.

A.REI.D.10

Understand that the graph of an equation in two variables is the set of all its solutions plotted in the coordinate plane, often forming a curve (which could be a line).

F.IF.C.7.A

Graph linear and quadratic functions and show intercepts, maxima, and minima.

F.LE.B.5

Interpret the parameters in a linear or exponential function in terms of a context.

What is Mathspace

About Mathspace