How to Solve September SAT Problems With Desmos

TL;DR

Use Desmos tables and regression to turn many SAT algebra, percentage, circle, and scale-factor questions into structured equations. The central method is to practice one topic repeatedly, translate every relationship carefully, and account for whether measurements scale linearly, by area, or by volume.

Transcript

I just took the September SAT a week early. This is what happened. So, I was supposed to take the August SAT in the Moscone Center in San Francisco, but so many people had Wi-Fi issues that like more than half of us couldn't even take the test. Going into the test, I kind of thought that it was going to be the same as like the August SAT is going t... Read More

Key Insights

• Desmos regression is a direct method for fitting a quadratic when three points are provided. Enter the coordinates in a table, select quadratic regression, rewrite the resulting equation in function notation, and then evaluate the requested expression, such as f(9) minus f(11).
• A general quadratic requires three independent values because its expanded form contains three constants. When only two points are available, Desmos cannot uniquely determine all three constants unless the problem supplies additional information, such as the coordinates of the vertex.
• Vertex form is useful when a quadratic problem provides a vertex and other points. Writing the model as y = a(x − h)^2 + k allows the known vertex coordinates to fix h and k, leaving regression to determine the remaining coefficient a.
• Percentage word problems can be translated into systems by assigning one variable to each quantity. Statements such as 53% of A equals 27% of B become equations, while an unknown requested percentage can be represented by a placeholder such as W.
• Repeated practice within one topic is the recommended study approach. The presenter suggests solving many percentage problems together, followed by many circle problems, so each recurring setup becomes familiar instead of appearing sporadically among unrelated question types.
• A point-on-circle condition can be checked visually in Desmos by graphing the circle and the stated point. Adjusting the unknown parameter until the circle passes through the point identifies the value that satisfies the geometric condition.
• The center of a circle is read with opposite signs from standard form because the coordinate terms are written as x − h and y − k. When diameter endpoints are involved, the midpoint formula connects the two endpoints to the known center.
• Scale factors depend on the type of measurement being compared. Side lengths use the linear factor, areas use its square, and volumes use its cube, so a volume ratio must be converted with cube roots before it is applied to an area relationship.

Questions & Answers

Q: How can Desmos solve a quadratic from three points?

Enter the three coordinate pairs into a Desmos table, then apply quadratic regression. Because a general quadratic contains three constants, the three points allow Desmos to determine its coefficients. Rewrite the resulting equation with function notation if required, then type the requested evaluations directly, such as f(9) − f(11). In the demonstrated problem, that expression produces 240.

Q: Why are two points insufficient for a general quadratic?

Two points do not provide enough information to determine the three constants in a general quadratic equation. A regression based only on those points therefore cannot identify one unique quadratic. The problem needs another constraint, such as a third point or a known vertex. With a known vertex, vertex form fixes two parameters and leaves the remaining coefficient to be calculated.

Q: How do you use vertex form with Desmos regression?

Write the quadratic as y = a(x − h)^2 + k, where h and k come from the stated vertex. Link the available table points to this expression and use regression to solve for a. In the example, the vertex information leads to h = 9 and k = −13, while regression gives a = 0.25. The completed function can then be transformed and evaluated.

Q: How can Desmos solve difficult SAT percentage problems?

Assign a variable to every quantity, then translate each percentage statement into an equation inside a system. For example, 53% of A equals 27% of B becomes one relationship. If the question asks what percent of B equals C, use W as the unknown percentage in a second relationship. Desmos regression can then solve for W, which is rounded as instructed.

Q: What does “was” mean in an SAT percentage equation?

In the demonstrated translation method, the word “was” signals an equality relationship. If the number of soccer players was 2,000% of the number of tennis players, assign variables to both groups and write the corresponding percentage equation. Additional statements, such as basketball players being 75% of soccer players, create the remaining equations needed to calculate the requested total.

Q: How can you find an unknown circle parameter in Desmos?

Type the circle equation into Desmos and plot the point that the question says lies on the circle. Adjust the unknown parameter until the graph passes through that point. The demonstrated circle reaches the point (9, 8) when the parameter s equals 97. Moving beyond that value causes the graph to stop passing through the point, confirming the fitted value.

Q: How do circle endpoints determine the center?

The center lies at the midpoint of two opposite endpoints, so average their x-coordinates and average their y-coordinates. Also read the center carefully from standard circle form, where the terms are x − h and y − k. In the example, the center coordinate is −7, and setting it equal to the midpoint expression (−15 + a) / 2 gives a = 1.

Q: How do area and volume scale factors differ?

A linear scale factor compares corresponding side lengths directly. Area changes according to the square of that factor, while volume changes according to its cube. Therefore, comparing volumes of 125 and 216 does not give the linear factor by simple division. Taking the cube roots produces the relevant scale factor, 1.2, which can then be squared for an area relationship.

Summary & Key Takeaways

• The September SAT included question types that differed from what the test taker expected based on August. The lesson reconstructs the tested topics without repeating the exact questions, beginning with easier Desmos techniques and progressing toward harder applications. The presenter predicts a 790 math score but remains uncertain about English.

• Quadratic problems can be approached with Desmos tables and regression. Three points determine the three constants in a general quadratic, while a problem with two points requires additional information, such as a known vertex. Vertex form incorporates that information before transformed functions and requested differences are evaluated.

• Percentage relationships become systems of equations by assigning variables to quantities and translating phrases such as “was” or “is equal to.” Circle problems use graphing, standard form, and the midpoint formula. Similar-figure problems require distinguishing side-length scale factors from squared area relationships and cubed volume relationships.