In our opinion Desmos will be a lot easier to use than TI- Inspire and it will save a lot of time, especially in Module 1 of the Maths section. However, it can be a bit tricky to use. Therefore, it will take time to get used to it.  We recommend you to try the question bank with the help of desmos in order to get used to it. 

Keep in mind you are still able to use the TI-nSpires, so use it if you prefer it. The nSolve function can be quite useful if you are used to using the calculator quickly. However, for graphing and regression Desmos is the recommended way to solve.

Solving Equations

Desmos can be used to find approximate solutions to any equation with a single variable. These equations will take on the form of f(x)=g(x). Consider the example, -4x2-7x=-36. In this case, you can use Desmos to graph both sides of the equation as follows:

Once you have both sides of the equation graphed, you can look for the intersection points. The x-values of these points represent solutions to the equation. The only issue with this method is that you will not always get an exact value solution since Desmos will only show decimals. If you would like an exact value, you can use Wolfram Alpha. 

Let’s try another example, 2|4-x|+3|4-x|=25. After graphing, we get the following:

Solving Systems of Equations

Desmos is quite effective at solving systems of equations. Graphically, solving a system of equations means finding the point(s) at which two curves intersect each other. In Desmos, the implementation intuitively follows from that point. You will graph the equations in the system, find where they intersect, and click on the points of intersection to find the ordered pairs that represent solutions to the system. Let’s do an example:

4x=20

-3x+y=-7

Finding Intercepts

Given a curve, it may be of interest to find the x and y intercepts, or the points where y and x equal 0, respectively. To do this, all you need to do is graph the curve and look for where it intersects. Desmos automatically marks intercept points, so all you need to do is click on these points to find the intercepts. Let’s try an example of this, where we find the intercepts of (-8)(2x)+22. 

As can be seen, the x-intercept is about (1.459, 0) and the y-intercept is (0, 14). This method also works for functions that contain multiple x-intercepts, with each point being one of the intercepts. 

Minimum and Maximum Values

Desmos is quite useful for solving minimum and maximum value problems from the SAT. It works quite well in general, but the SAT generally only asks questions about quadratics, so all you need to do to solve these problems is report the coordinates of the vertex.This means that all you need to do is locate this point. Let’s try an example of this, with the equation x2-14x+22.

Once you locate the vertex and click on it, you can see the x and y coordinates. The y coordinate will always represent either a minimum or maximum of this graph. 

Inequalities

Desmos can be used to check if a particular point or set of points represent a solution to an inequality. To do this, there is a two-step process:

1. Graph the inequality

2. Check each point to see if it is a solution

Let’s consider the inequality, y<6x+2 with the points (3, 16), (5, 28), and (7, 40). First, we need to graph inequality, which will give this graph:

From here, we need to create the points. Desmos has a table that allows you to graph many points at once, which is very helpful in this case. To create a table, press the plus sign on the top left and then look for the menu item titled “table”. Use this to create the 

Since all the points are in the shaded region, they represent valid solutions to the system. If the points were not shaded, they would not be solutions to the system. Additionally, if they are exactly on the boundary of the shaded region, they are solutions if the line is solid (there is a or sign) and they are not solutions if the line is dotted (there is a < or > sign). 

It is also possible to use this method for multiple inequalities. To do this, graph the second inequality and then look at the region shaded by both. Points that are shaded by both inequalities are solutions to both inequalities. An example is below, with the inequality y>-x being added to the previous example. 

How Many Solutions?

Some questions in the SAT ask how many solutions a particular system of equations has. To do this, all you need to do is look at the graph of the system and check how many times they intersect. That is why the term “points of intersection” is used interchangeably with “solutions” in these types of problems. 

In this case, there is one solution since the lines intersect once. If the lines overlap, there are infinitely many solutions. If the lines are parallel but not overlapping, then there is no solution. There are the only options for systems with only lines. If other curves are included, it is possible to have a different number of solutions. 

How Many Solutions? With constants

Another type of question that the SAT asks is to find the value of a constant such that there are either infinitely many or 0 solutions. The most efficient way to do this is to use sliders to find guess-and-check. One example of such a question is as follows,

4x-9y=9y+5

hy=2+4x

Where you are supposed to find the value of h such that there is no solution to the system. This can be done as follows,

Desmos will have a popup button that allows you to create a slider for the equation in the bottom. Adjust the value. You will notice that no value on the slider will work. To solve this problem, extend the range of the slider by clicking the number 10 on the right. 

Once you extend the range and slide again, you will find the solution to be h=18. There are other similar scenarios in which sliders allow you to guess and check solutions much faster than using arithmetic. 

Evaluating Functions

Desmos allows you to graph and evaluate a function very conveniently. Consider the function f(x)=ex+4x2. Once you have it written in Desmos, you can evaluate the function by writing f(a specific number). For example, the graph below shows the function being evaluated at x=3, giving a result of about 56.

Regression

Normally, linear regression involves finding values for the slope and y-intercept for a line that closely matches a set of data points. The SAT simplifies this by asking you to find the line that perfectly matches three points. Desmos has a built-in function that is able to do this. First, insert a table in the same way as covered in the inequalities section. Once you get a graph like below for the set of data points, you are ready to move on. 

Once you have obtained the graph, you can write the relationship between y and x, which is a line in slope-intercept form. However, use ~ instead of = when writing this so that Desmos knows to do a regression. Once you have this, you will be able to see the value of m, the slope, as well as b, the y-intercept. 

This can also be applied for quadratic functions with three points, using y1~ax12+bx1+c instead of a line. This equation represents a parabola and will give you values of a, b, and c that match the three points. Given a set of points and any function to be mapped onto the points, the same technique applies. For example, you could try an exponential function. 

Equivalent expressions

There are some questions in the SAT that ask you to select one of 4 expressions that is equal to some expression stated in the question. These can be solved by graphing the expression in the question and then checking to see which answer choice has a graph that entirely overlaps with the expression in the question. 

Equations of Circles

There are some questions on the SAT that give you the equation of a circle and ask you to find the midpoint, radius, or diameter. To solve these problems, you can use the built-in functions for midpoint or distance. All you need to do is graph the circle, select the points that automatically appear when you click on the circle, and then use the coordinates of two opposite points to calculate the midpoint and diameter. This can be done as suggested in the below diagram. 

While these questions are simple, there are a few more complicated cases where you need to insert a number into a list that will cause the mean and median to obey certain conditions. The best way to do this is to graph a function of the variable inserted, as follows:

If you were asked to find the values of x such that the mean and the median are the same, you could read off the intersection points of this graph to do so. By using Desmos, you will save time from arithmetic and minimize the chances of making mistakes. 

Sliders

As mentioned previously in “How Many Solutions? With constants”, sliders can be used for efficient guessing and checking. Refer to that section for a tutorial on how to use them. While systems of equations are the main use-case, you may wish to use them for other problems that you do not want to solve algebraically. 

Solving Equations Quickly

As mentioned in the section “Solving Equations”, you can graph both sides of the equations, find the intersection points, and then report the x-values as solutions. However, there is a faster way, which is to just directly type the equation into desmos. Once you do this, you will have vertical lines appear at the correct values as follows:

You may then click on these lines to find the values. Sometimes, this method can be buggy; if it does not work as intended, refer to the “Solving Equations” section to solve the equation. 

Percents

For most percent problems, the challenge is setting up the algebra, so Desmos functions most effectively as a calculator. Remember that to convert a percent to a fraction, divide by 100. Also remember that the percentage of a quantity is the same as multiplying the quantity by the corresponding fraction. For example, 25% of 50 =50*(25/100)=12.5 . 

Another important concept is the percent multiplier, which is either reported as an increase or a decrease. With percent p, this multiplier is equal to 1+p100 for increases and 1-p100 for decreases. The percent change is equal to 100(PM-1) where PM is the percent multiplier. Questions in the math section of the SAT may make reference to either percent multipliers or percent changes. You may also calculate this directly, using the formula, new-oldold.

Domains

Given a particular problem, you may feel that it is appropriate to restrict the domain. To do this, you can use the syntax, y=f(x){a<x<b}, where a and b are the bounds of the domain. Either one of them may be infinite. An example is shown below.

Tips for Student Product Responses

The SAT math section includes student produced responses. On these questions, you will type a number into a box. This may be an integer, decimal, or fraction Here are some tips for filling out these types of responses:

1. Do not enter symbols such as “%”, “,”, and “$”

2. Use improper fractions or decimals instead of mixed numbers

3. Use decimals if fractions do not fit and vice versa

4. If necessary, truncate or round responses to 5 characters. Only do this for decimal responses. 

5. Since Desmos always rounds to the thousands place, it is accurate enough for these questions

Parameterization

For certain problems, such as finding parametric curves that do not intersect a particular line, sliders can be used. However, parameterization is a trick that can solve the problem much faster. It requires using the notation to graph a point, but instead of inputting numbers you input functions. The expression will look like (f(t), g(t)). It is important that you use t as the dependent variable, since desmos is programmed to recognize it for this purpose. You may also specify the domain of t using the input boxes below the expression once you type it. An example is below.