Skip to main content
BuildSAT.

FREE SAT STUDY GUIDE

SAT Desmos Strategies

Learn when and how to use Desmos for SAT math, with graphing walkthroughs, worked examples, and common mistakes.

These are BuildSAT teaching examples. Use the study workspace to apply the concepts to official practice questions and review your progress.

Open the Desmos lab →

See the whole problem

A graph looks empty, intersections seem missing, or the values are much larger than the default window.

  1. Enter both equations. The default window can hide the intersection entirely.y=0.02x+100
  2. Enter the second line.y=-0.03x+350
  3. Open graph settings. Set x from 0 to 6000 and y from 0 to 400. Tap the intersection: (5000, 200). Equal axis scales are not necessary for solving a system.
  4. Check the requested quantity by substituting x = 5000 into either original equation. Both give 200.

Try an example

Try it: y = 0.5x + 1200 and y = 0.2x + 1800. Find x.

Show explanation

x = 2000, y = 2200. Use an x-window of 0 to 3000 and a y-window of 1000 to 3000. Algebra: 0.3x = 600.

Strategy: A blank screen is not evidence of no solutions. Choose a window using the quantities in the problem.

Common mistakes

  • A rounded coordinate can hide the exact fraction. Substitute or solve algebraically before entering an exact answer.
  • An apparent crossing is not precise enough; tap the intersection.

Test answer choices efficiently

A multiple-choice question asks which expression matches a given one.

  1. Define the original expression as a function.f(x)=(x+3)^2-x^2
  2. Define a candidate answer and compare values at more than one ordinary x-value.g(x)=6x+9
  3. Graph the difference. It lies on y = 0 wherever both expressions are defined.y=f(x)-g(x)
  4. Confirm with expansion: x² + 6x + 9 − x² = 6x + 9. Numerical agreement is a screening strategy, not a proof of identity.

Try an example

Try it: is (x² − 9)/(x − 3) equivalent to x + 3 for every real x?

Show explanation

They agree wherever the original expression is defined, but x = 3 is excluded. Canceling a factor does not restore an excluded input. Desmos may make the missing point hard to see.

Strategy: Avoid only testing x = 0 or 1; different expressions often coincide at those convenient values.

Common mistakes

  • A shared graph at normal zoom does not prove exact equivalence.
  • Check denominators, square-root domains and restrictions before choosing an answer.

Solve for an unknown constant

An equation has one solution, no solutions, or depends on a parameter.

  1. Define a parameter and graph the expression. Desmos will offer a slider for k.y=x^2-6x+k
  2. Move k until the vertex touches the x-axis. The boundary occurs at k = 9; higher k gives no real zeros and lower k gives two.
  3. Confirm the exact boundary by completing the square: x² − 6x + k = (x − 3)² + k − 9. Tangency requires k − 9 = 0.
  4. For two linear equations with infinitely many solutions, compare both slopes AND intercepts. A slider alone cannot establish an exact identity.

Try an example

Try it: x² − 8x + c = 0 has exactly one real solution. Find c.

Show explanation

c = 16. Completing the square gives (x − 4)² + c − 16 = 0. Equivalently the discriminant 64 − 4c must equal 0. The graph touches the axis at x = 4.

Strategy: Use sliders to understand the shape, then algebra to pin down an exact constant.

Common mistakes

  • A slider step of 1 can miss fractional parameter values.
  • One visible intersection is not proof that another is absent outside the window.

Read overlapping solution regions

You must identify a point satisfying several constraints.

  1. Graph the first inequality. Its dashed boundary is excluded.y>2x-1
  2. Graph the second. Its solid boundary is included.y<=-x+5
  3. The overlap contains points satisfying BOTH. Test (1, 3): 3 > 1 and 3 ≤ 4, so it works.
  4. For a real-world nonnegative input, restrict the graph.y=2x-1{x>=0}

Try an example

Try it: does (2, 3) satisfy y > 2x − 1 and y ≤ −x + 5?

Show explanation

No. The first condition becomes 3 > 3, which is false. The second is true, but a system requires every condition. Boundary inclusion matters.

Strategy: Shading narrows the search. Substitute the chosen point into every original inequality.

Common mistakes

  • A strict inequality excludes its boundary.
  • Do not round a point onto a boundary and assume it remains a solution.

Solve nonlinear systems

A system includes a quadratic, circle, or other nonlinear relation.

  1. Enter the quadratic.y=x^2
  2. Enter the line and inspect BOTH intersections.y=2x+3
  3. The intersections are (−1, 1) and (3, 9). If x must be positive, only (3, 9) qualifies.
  4. If asked for the sum of all possible x-values, compute −1 + 3 = 2, not the sum of y-values.

Try an example

Try it: y = x² and y = 4. Find the sum of the possible x-values.

Show explanation

The intersections are (−2, 4) and (2, 4), so the sum is 0. Both solutions matter even though their y-values are identical.

Strategy: Translate the final question into a coordinate or combination before you graph.

Common mistakes

  • A nonlinear system can have more than one solution.
  • Check positivity, integrality, and context restrictions afterward.

Model growth and decay

A quantity changes by a constant percent or repeated factor.

  1. A quantity starts at 500 and loses 20% per hour. Define the function with multiplier 0.8.f(x)=500(0.8)^x
  2. To find when the quantity reaches 256, graph the target level.y=256
  3. The intersection is at x = 3. Check 500 × 0.8³ = 256.
  4. If a table follows an exponential model, use a table and y_1 ~ a b^(x_1). Verify the units: x measured in hours is different from x measured in half-hours.

Try an example

Try it: an amount of 200 grows by 10% each year. Find the amount after 3 years.

Show explanation

200(1.1)³ = 266.2. The increase is 66.2, whereas the final amount is 266.2. Read which quantity is requested.

Strategy: The multiplier is 1 + rate for growth and 1 − rate for decay, with the rate written as a decimal.

Common mistakes

  • 20% decay means multiply by 0.8, not 0.2.
  • If a multiplier applies every 5 years, the exponent is t/5 when t counts years.

Translate percent and rate problems

Arithmetic is easy after choosing the right base quantity and units.

  1. An original price x increases 25%, then decreases 20%. Enter the combined multiplier.1.25*0.8
  2. The product is 1, so this particular pair of changes returns to the starting price. Equal percentage increases and decreases generally do not cancel.
  3. To reverse a 15% discount from a final price of 68, divide by the remaining fraction.68/0.85
  4. The original price is 80. For unit conversions, write canceling units on paper before entering the arithmetic.

Try an example

Try it: a speed is 90 kilometers per hour. Convert to meters per second.

Show explanation

90 × 1000 / 3600 = 25 meters per second. The kilometers cancel with 1000 meters/kilometer and hours cancel with 3600 seconds/hour.

Strategy: Desmos can calculate the expression; you must decide what the percentage is of.

Common mistakes

  • A change from 40 to 50 is a 25% increase, not 10%.
  • A percent of a percent multiplies the two decimal rates.

Lists, center and spread

You have a small data list, frequency table, or a transformation of every observation.

  1. Enter the list with square brackets.L=[2,4,4,6,9]
  2. Compute mean(L) = 5 and median(L) = 4. Do not enter distinct values only when values repeat.mean(L)
  3. To compare population standard deviations, use stdevp(L). stdev(L) uses the sample formula. Choose the convention the question specifies.stdevp(L)
  4. Adding 10 to every observation shifts mean and median by 10, but leaves standard deviation unchanged. Multiplying every observation by 3 multiplies standard deviation by 3.

Try an example

Try it: compare the standard deviations of [1,3,5] and [11,13,15].

Show explanation

They are equal: the second list adds 10 to every observation, preserving every distance from the mean. Their means are 3 and 13. Population standard deviation is √(8/3) for each.

Strategy: When the question asks only how spread changes, reasoning about the transformation is often faster than calculating.

Common mistakes

  • A frequency of 7 means the value occurs seven times.
  • A calculator cannot decide whether a sample is representative or an experiment supports causation.

Evaluate and transform functions

You need f(a), a shifted graph, or an input that produces a given output.

  1. Define a reusable function.f(x)=x^2-4x+1
  2. Enter f(3) to evaluate: the result is −2.f(3)
  3. Now define g(x) = f(x − 2) + 5. The graph shifts RIGHT 2 and UP 5.g(x)=f(x-2)+5
  4. The original vertex is (2, −3); the transformed vertex is (4, 2). A shift inside the input has the opposite sign from the direction.

Try an example

Try it: f(x) = (x − 1)² + 3 and g(x) = f(x + 4) − 2. Find the vertex of g.

Show explanation

The original vertex (1, 3) shifts left 4 and down 2, giving (−3, 1). Expanding the definition gives g(x) = (x + 3)² + 1.

Strategy: Keep input, output and function name separate. f(3) is an output; solving f(x) = 3 asks for input(s).

Common mistakes

  • f(x) + k changes outputs; f(x + k) changes the horizontal position.
  • A restricted domain can remove a vertex from the allowed part of a graph.

Recover coefficients from points

A quadratic is known to pass through three distinct input-output pairs.

  1. Create a table: x₁ = 0, 1, 2 and y₁ = 3, 6, 13. Each row is one point.
  2. Fit a quadratic using a tilde.y_1~ax_1^2+bx_1+c
  3. Read a = 2, b = 1, c = 3. If asked for a + b + c, the answer is 6; the question may not ask for the entire equation.
  4. Verify all three points in y = 2x² + x + 3. Three distinct x-values determine a quadratic of degree at most 2; repeated x-values do not add the same information.

Try an example

Try it: a quadratic passes through (0, 1), (1, 4), (2, 9). Find a, b and c in ax² + bx + c.

Show explanation

a = 1, b = 2, c = 1. At x = 0, c = 1; at x = 1, a + b = 3; at x = 2, 4a + 2b = 8. Solve to verify the regression.

Strategy: Regression is an efficient way to recover numeric coefficients from exact data. For symbolic conditions, build equations and verify analytically.

Common mistakes

  • A near-zero residual does not prove that rounded coefficients are exact.
  • Too few independent points leave multiple possible models.

Check domains and extra roots

Squaring, denominators, or domain restrictions may create invalid candidate solutions.

  1. To solve √(x + 2) = x, graph the two original sides.y=sqrt(x+2)
  2. Enter y = x. The only intersection is (2, 2).y=x
  3. Squaring gives x² − x − 2 = 0 and candidate values −1 and 2. Substitute into the ORIGINAL equation: √1 = 1, which is not −1. Reject −1.
  4. For rational equations, record where every denominator is zero before canceling factors or multiplying.

Try an example

Try it: solve (x² − 4)/(x − 2) = 4.

Show explanation

There is no solution. Simplifying for x ≠ 2 gives x + 2 = 4 and candidate x = 2, but x = 2 makes the original denominator zero. A graph may visually hide this hole.

Strategy: Graph the original expressions as a check, but write domain restrictions explicitly.

Common mistakes

  • Squaring both sides can introduce extra solutions.
  • Canceling a common factor never makes its excluded input legal.

Turn geometry into coordinates

A diagram provides equations or measurements you can translate into coordinates and formulas.

  1. A circle centered at (2, −1) with radius 5 has this equation.(x-2)^2+(y+1)^2=25
  2. Intersect it with y = 3. The x-coordinates are −1 and 5; the horizontal chord length is 5 − (−1) = 6.y=3
  3. For similar figures, a length scale factor of 3 produces an area factor of 9 and a volume factor of 27. Compute the appropriate power.
  4. For an inverse trig angle with opposite 6 and adjacent 8, use arctan(6/8) in degrees mode: about 36.87°. Use only the diagram’s stated constraints.

Try an example

Try it: a circle of radius 10 has a chord 6 units from its center. Find the chord length.

Show explanation

Half the chord is √(10² − 6²) = 8, so the full chord is 16. You can graph x² + y² = 100 and y = 6; intersections at x = −8 and 8 confirm the result.

Strategy: Choose the geometric relationship first, then let Desmos handle arithmetic and intersections.

Common mistakes

  • A radius, diameter, circumference and area are different requested quantities.
  • For arc length rθ and sector area r²θ/2, θ must be in radians.

Put the ideas into practice

Choose this domain in the question bank, review the explanation after each attempt, and use a mastery check to find what to revisit.

Open practice questions →Check your mastery

Explore related guides

SAT AlgebraSAT Advanced MathSAT Problem-Solving and Data AnalysisSAT Geometry and TrigonometrySAT Information and IdeasSAT Craft and StructureSAT Expression of IdeasSAT Standard English Conventions