HSA.REI.D.11: Solving f(x) = g(x) Using Graphs, Tables and Approximations
In plain English: HSA.REI.D.11 is the Common Core algebra standard that asks students to explain why the x-coordinates of the intersection points of y = f(x) and y = g(x) are the solutions of f(x) = g(x), and to find them approximately with graphing technology, tables or successive approximations. It covers linear, polynomial, rational, absolute value, exponential and logarithmic functions and usually spans Algebra I and II.
Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the solutions approximately, e.g., using technology to graph the functions, make tables of values, or find successive approximations. Include cases where f(x) and/or g(x) are linear, polynomial, rational, absolute value, exponential, and logarithmic functions.
Common Core State Standards for Mathematics · Domain: Reasoning with Equations and Inequalities (REI) · Cluster: Represent and solve equations and inequalities graphically Also written as HSA-REI.D.11 or A-REI.11 · Official standard
In this lesson, students learn why the equation f(x) = g(x) can be solved by graphing y = f(x) and y = g(x) on the same axes: an intersection point (a, b) lies on both graphs, so b = f(a) and b = g(a), and therefore f(a) = g(a). The x-coordinate a is a solution, and the y-coordinate b is the common value of the two sides.
Students then find solutions approximately in three ways named in the standard: with technology (graphing and the intersect feature), with tables of values that show where f(x) - g(x) changes sign, and with successive approximations that narrow an interval step by step. The examples cover every function type the standard lists: linear, polynomial, rational, absolute value, exponential and logarithmic. Many of these equations, such as 2ˣ = x + 3, cannot be solved exactly with algebra taught in high school, which is why the approximate methods matter.
Learning Objectives
By the end of this lesson, students will be able to:
Explain why the x-coordinates of the intersection points of y = f(x) and y = g(x) are the solutions of f(x) = g(x)
Use technology to graph two functions and approximate the x-coordinates of their intersections
Use a table of values and a change in sign of f(x) - g(x) to locate a solution between two inputs
Refine a solution by successive approximations to a required precision, such as the nearest hundredth
Solve equations where f and g are linear, polynomial, rational, absolute value, exponential or logarithmic, and check each solution
Prior Knowledge Required
Students should already be comfortable with:
Knowing that a graph is the set of all solutions of its equation HSA.REI.D.10
Solving systems of equations by graphing HSA.REI.C.6
Using function notation and evaluating functions for given inputs HSF.IF.A.2
Recognizing graphs of linear, quadratic, absolute value and exponential functions HSF.IF.C.7
Evaluating logarithms such as log₂ 8 = 3; review this briefly before the logarithmic example if needed
Display the prompt. Students work alone for 3 minutes, then compare with a partner.
Warm-Up Prompt
"Try to solve 2ˣ = x + 3. Test some values of x in a table. Can you find an exact answer? Can you find at least one answer to the nearest whole number?"
Students usually find that x = 2 gives 4 versus 5 and x = 3 gives 8 versus 6, so a solution lies between 2 and 3. Few will look for the negative solution. Ask: "Is there an algebra step that isolates x here?" There is not with the tools of this course, which motivates the graphical and numerical methods of the lesson.
Direct Instruction20-25 minutes
Start with the reason, then the methods:
Why it works: if (a, b) is on the graph of y = f(x), then b = f(a). If it is also on the graph of y = g(x), then b = g(a). So f(a) = g(a), and a is a solution. Conversely, if f(a) = g(a), the point (a, f(a)) is on both graphs.
Technology: graph y₁ = f(x) and y₂ = g(x), find every intersection (widen the window to check for more), and record only the x-coordinates.
Tables: list f(x) and g(x) side by side. Where f(x) - g(x) changes sign between two inputs, and both functions are continuous there, a solution lies between them.
Successive approximations: split the interval into tenths, find where the sign changes, then split that interval into hundredths, and repeat until you reach the required precision.
Check: substitute each answer into both sides; with approximate answers the two sides should be very close.
Work these examples, one or more for each function type named in the standard:
Absolute value and linear
Solve |x - 2| = 0.5x + 1 by graphing y = |x - 2| and y = 0.5x + 1. The graphs cross twice. Confirm with the two cases x - 2 = 0.5x + 1 and -(x - 2) = 0.5x + 1.
Equation: x = 2/3 and x = 6 (intersections (2/3, 4/3) and (6, 4))
Exponential and linear
Solve 2ˣ = x + 3. A table shows 2² = 4 < 5 but 2³ = 8 > 6, so a solution lies between 2 and 3. It also shows 2⁻³ = 0.125 > 0 but 2⁻² = 0.25 < 1, so a second solution lies between -3 and -2. Technology refines both.
Equation: x ≈ -2.86 and x ≈ 2.44
Polynomial and linear
Solve x³ = x + 1 by successive approximations. x³ - (x + 1) changes sign between 1.3 and 1.4, then between 1.32 and 1.33, then between 1.324 and 1.325.
Equation: x ≈ 1.32 (to the nearest hundredth)
Rational and linear
Solve 6/x = x + 1 by graphing y = 6/x and y = x + 1. Read the intersections, then check: multiplying by x (x ≠ 0) gives x² + x - 6 = 0.
Equation: x = 2 and x = -3 (intersections (2, 3) and (-3, -2))
Logarithmic and linear
Solve log₂ x = x - 2. The table value x = 4 gives log₂ 4 = 2 and 4 - 2 = 2 exactly. The graph shows a second intersection close to x = 0.3, which technology refines.
Equation: x = 4 and x ≈ 0.31
Guided Practice15 minutes
Pairs solve three equations with the method named on each card: 8/x = x - 2 by graphing (x = 4 and x = -2), 2ˣ = x² by technology (x = 2, x = 4 and x ≈ -0.77), and x³ = 3 by successive approximations (x ≈ 1.44). For each, students write one sentence that names the functions f and g, and one sentence that explains why the answer is an x-coordinate. Watch for students who report the intersection point instead of the x-coordinate, and for students who miss an intersection outside the default calculator window, which is common with 2ˣ = x².
Independent Practice15-20 minutes
Students solve four equations on their own, one each with a rational, an absolute value, an exponential and a logarithmic function paired with a linear function. At least one must be done with a table and one with successive approximations to the nearest tenth. For each, students record the method, the answer and a check by substitution into both sides.
Closure5-10 minutes
Exit ticket: "The graphs of y = f(x) and y = g(x) intersect at (-1, 4) and (3, 0). (a) What are the solutions of f(x) = g(x)? (b) Explain in one or two sentences why 4 and 0 are not solutions." Look for x = -1 and x = 3, and for the reason that the y-coordinates are the shared output values, not inputs that make the equation true.
Differentiation Strategies
For Struggling Students
Give a three-column table template (x, f(x), g(x)) plus a fourth column for "which is bigger?" so the sign change is easy to spot
Start with equations that have integer solutions, such as log₂ x = 6 - x (x = 4) or 6/x = x + 1, before moving to approximate ones
Provide a calculator keystroke card for graphing two functions and using the intersect feature
For Advanced Students
Explain why a sign change in f(x) - g(x) does not guarantee a solution for 1/x = 0.5 on the interval from -1 to 1, and what goes wrong
Find how many solutions 2ˣ = x² has and prove there are no more than the ones found, using the growth of 2ˣ compared with x²
Write an equation f(x) = g(x) with a logarithmic f that has exactly one solution, and justify it with a graph
Assessment Guidance
What to Look For
Listen for the reason, not just the answer: students should say that an intersection point satisfies both equations, so its x-value makes f(x) and g(x) equal. Check that students give x-values rather than points, search for all intersections, state the precision of approximate answers, and check solutions of absolute value and rational equations, where algebraic steps can produce values that are not solutions.
02
Classroom Activities
3 Activities
1
Function Pair Stations
25 minGroups of 3-4
Set up six stations, one for each function type in the standard. At each station, groups solve one equation f(x) = g(x) with the method on the card and record the answer to the nearest hundredth.
Station Cards
Linear: 0.5x + 3 = -2x + 8 (graphing; x = 2)
Polynomial: x³ - 4x = 1 (technology; x ≈ -1.86, x ≈ -0.25 and x ≈ 2.11)
Rational: 12/x = x + 4 (graphing; x = 2 and x = -6)
Absolute value: |2x - 1| = x + 4 (graphing; x = -1 and x = 5)
Exponential: 3ˣ = 10 - x (table, then successive approximations; x ≈ 1.90)
Logarithmic: ln x = x - 2 (technology; x ≈ 0.16 and x ≈ 3.15)
Procedure
Groups spend about 4 minutes per station and rotate on a signal
At each station they sketch both graphs, circle the intersection points and write the solutions as x-values
Before leaving a station, one member checks each answer by substituting into both sides
Modification for Distance Learning
Post each station as a slide in a shared deck with a link to a graphing app; groups paste a screenshot of their intersections and type their check.
2
Zoom In: Successive Approximation Race
15-20 minPairs
Pairs use only a scientific calculator (no graphing) to find the positive solution of x³ - 2x = 5 to the nearest hundredth, recording every interval they test.
Procedure
Step 1: test whole numbers; 2³ - 4 = 4 < 5 and 3³ - 6 = 21 > 5, so the solution is between 2 and 3
Step 2: test tenths; 2.0 gives 4 and 2.1 gives 5.061, so the solution is between 2.0 and 2.1
Step 3: test hundredths from 2.00 to 2.10; 2.09 gives about 4.9494 and 2.10 gives 5.061, so the solution is between 2.09 and 2.10
Step 4: test 2.095 (about 5.005, just over 5) to decide the rounding: the solution is between 2.09 and 2.095, so x ≈ 2.09
Discussion Questions
Why is testing 2.095 needed to round correctly to the nearest hundredth?
How many more steps would you need for the nearest thousandth?
Check your answer with a graphing tool. How close were you?
3
Which Town Grows Faster?
20 minIndividual then share
Town A has 5,000 residents and grows 4% per year: A(t) = 5000(1.04)ᵗ. Town B has 7,000 residents and adds 100 residents per year: B(t) = 7000 + 100t. Students find when the towns have equal populations by solving A(t) = B(t).
Tasks
Make a table for t = 0, 5, 10, 12, 13, 15 and find where A(t) - B(t) changes sign (at t = 12, A ≈ 8,005 and B = 8,200; at t = 13, A ≈ 8,325 and B = 8,300)
Graph both functions with technology and use the intersection to refine the answer (t ≈ 12.9 years)
Explain what the x-coordinate and the y-coordinate of the intersection mean for the towns
Extension Variation
Ask whether the graphs intersect again for t > 13. Students test large values, such as t = 50, and argue that exponential growth stays ahead of linear growth, so there is only one positive solution. This is a model, so the populations are estimates, not exact counts.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Solving 2ˣ = x + 3 as an Intersection of Two Graphs
Drawn to scale. The graphs of y = 2ˣ and y = x + 3 cross at two points. Their x-coordinates, about -2.86 and 2.44, are the solutions of 2ˣ = x + 3. The y-coordinates are the shared values of both sides, not solutions.
Diagram 2: Narrowing In on the Solution of x³ = x + 1
Each step zooms in on the interval where x³ - (x + 1) changes from negative to positive. The green dot marks the actual solution, x ≈ 1.3247, at its true position on each number line.
04
Homework Assignment
~30 min
HSA.REI.D.11 Homework: Where the Graphs Meet
Directions: For each problem, name the functions f and g, state the method you used (graph, technology, table or successive approximations), and give every solution as an x-value. Round approximate answers as instructed and check each answer by substituting into both sides.
Part 1: Exact Intersections (Problems 1-3)
Solve |x + 1| = 2x - 4 by graphing y = |x + 1| and y = 2x - 4. Then explain why solving the two cases algebraically produces x = 1, which is not a solution.
Solve 4/(x - 1) = x + 2 by graphing both sides. Give the intersection points and the solutions, and explain why x = 1 cannot be a solution.
The graphs of f(x) = x² - 2x and g(x) = x + 4 intersect at (-1, 3) and (4, 8). What are the solutions of x² - 2x = x + 4? Verify each one, and explain why 3 and 8 are not solutions.
Part 2: Approximate Solutions (Problems 4-6)
Use a table of values to find the positive solution of x³ - 3x = 4 to the nearest tenth. Show the table rows that locate the solution.
City A's population is modeled by A(t) = 6000(1.05)ᵗ and City B's by B(t) = 9000 + 100t, where t is years from now. Use a table or technology to find when the populations are equal, to the nearest tenth of a year. At the end of which year does City A first have more residents than City B?
Show that x = 3 is a solution of log₃ x = x - 2. Then use a graph or successive approximations to find the other solution to the nearest hundredth.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Setup
f and g named correctly and graphed or tabulated
One function set up incorrectly
No setup shown
Solutions
All solutions found, given as x-values with correct precision
One solution missing, or answers given as points
Solutions incorrect or missing
Method
Table, graph or approximation steps clearly shown
Method shown but incomplete
No method shown
Check and Explanation
Answers checked in both sides; reasoning about intersections is correct
Check or explanation incomplete
Missing
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
The graphs of y = x² - 3 and y = 2x intersect at (-1, -2) and (3, 6). What are the solutions of x² - 3 = 2x?
Answer: C
The solutions are the x-coordinates of the intersection points: x = -1 and x = 3. Check: 1 - 3 = -2 = 2(-1), and 9 - 3 = 6 = 2(3). Choice A lists the y-coordinates. Choice D mixes a y-coordinate with an x-coordinate.
Question 2 of 20 · Multiple Choice
The graphs of y = f(x) and y = g(x) intersect at the point (2, 5). Which statement must be true?
Answer: B
The point (2, 5) is on both graphs, so f(2) = 5 and g(2) = 5. The two sides are equal when x = 2, so x = 2 solves f(x) = g(x). Choice A uses the y-coordinate as the solution. Choice D confuses intersections of two graphs with x-intercepts.
Question 3 of 20 · Multiple Choice
Using successive approximations for x³ = 6, a student finds 1.81³ ≈ 5.930 and 1.82³ ≈ 6.029. Between which values is the solution?
Answer: D
x³ is less than 6 at 1.81 and greater than 6 at 1.82, so the solution lies between them (it is about 1.817). Choice A is below both values, where x³ is still less than 6 (1.80³ = 5.832). Choice B is past the sign change, where x³ is already greater than 6.
Question 4 of 20 · Multiple Choice
A table shows f(x) = 2x + 3 and g(x) = 2ˣ. x = 2: f = 7, g = 4 x = 3: f = 9, g = 8 x = 4: f = 11, g = 16 Where is the positive solution of 2x + 3 = 2ˣ?
Answer: A
f(x) - g(x) is positive at x = 3 (9 - 8 = 1) and negative at x = 4 (11 - 16 = -5), so the graphs cross between 3 and 4 (the solution is about 3.25). Choice C picks the closest row, but 9 ≠ 8. Choice D ignores that the solution can fall between table values.
Question 5 of 20 · Multiple Choice
The graphs of y = |x - 3| and y = 2 are drawn on the same axes. What are the solutions of |x - 3| = 2?
Answer: A
The horizontal line y = 2 crosses the V-shaped graph at x = 1 and x = 5: |1 - 3| = 2 and |5 - 3| = 2. Choice B misses the left branch of the V. Choice C comes from solving x - 3 = -2 incorrectly as x = -1.
Question 6 of 20 · Multiple Choice
At which x-value do the graphs of y = log₂ x and y = -x + 3 intersect?
Answer: C
At x = 2, log₂ 2 = 1 and -2 + 3 = 1, so the graphs meet at (2, 1). Choice A gives log₂ 1 = 0 but -1 + 3 = 2. Choice B confuses the line's y-intercept with a solution. Choice D gives 2 and -1.
Question 7 of 20 · Multiple Choice
Graph y = 10/x and y = x - 3. What are the solutions of 10/x = x - 3?
Answer: B
The graphs cross at (5, 2) and (-2, -5). Check: 10/5 = 2 = 5 - 3, and 10/(-2) = -5 = -2 - 3. Multiplying by x gives x² - 3x - 10 = 0, which confirms both. Choice A has the signs reversed from factoring as (x + 5)(x - 2). Choice C misses the intersection on the left branch. Choice D includes x = 0, where 10/x is undefined.
Question 8 of 20 · Multiple Choice
How many real solutions does 2ˣ = x² + 1 have?
Answer: D
x = 0 and x = 1 both work (1 = 1 and 2 = 2), and a third solution lies between 4 and 5, since 2⁴ = 16 < 17 = 4² + 1 while 2⁵ = 32 > 26 = 5² + 1. Technology gives x ≈ 4.26. For x < 0, 2ˣ < 1 < x² + 1, so there are no negative solutions. Choice A reflects a common mistake: testing only small values or using a narrow graphing window.
Question 9 of 20 · Multiple Choice
To solve log x = 0.5x - 1 graphically, what should you do?
Answer: B
Treat each side as a function, f(x) = log x and g(x) = 0.5x - 1. Where the graphs intersect, the two sides are equal, and the x-coordinates are the solutions. Choice A solves log x = 0 instead. Choice C reports the shared output values instead of the inputs.
Question 10 of 20 · Multiple Choice
A calculator graph of y₁ = x³ - 4x and y₂ = 2 shows intersections at x ≈ -1.675, x ≈ -0.539 and x ≈ 2.214. What are the solutions of x³ - 4x = 2?
Answer: A
Each intersection's x-coordinate makes x³ - 4x equal to 2, so all three are solutions. Choice B gives the shared output value, not an input. Choice C gives the zeros of x³ - 4x, which solve x³ - 4x = 0, not x³ - 4x = 2. Choice D keeps only the positive solution.
Question 11 of 20 · Multiple Choice
How many solutions does |x| = 2x + 3 have?
Answer: D
For x ≥ 0, x = 2x + 3 gives x = -3, which is not in that case, so it is rejected. For x < 0, -x = 2x + 3 gives x = -1, and |-1| = 1 = 2(-1) + 3. The graphs meet only once. Choice A keeps x = -3 without checking: |-3| = 3 but 2(-3) + 3 = -3.
Question 12 of 20 · Multiple Choice
The graphs of f(x) = 3x - 7 and g(x) = -x + 5 intersect at one point. What is the solution of 3x - 7 = -x + 5?
Answer: C
4x = 12, so x = 3; the intersection point is (3, 2), since 3(3) - 7 = 2 and -3 + 5 = 2. Choice A gives the y-coordinate of the intersection. Choice D stops at 4x = 12 and reports 12.
Question 13 of 20 · Multiple Choice
Account 1 grows as V = 500(1.06)ᵗ and account 2 as V = 600 + 20t, in dollars after t years. t = 5: $669.11 and $700 t = 6: $709.26 and $720 t = 7: $751.82 and $740 Comparing the balances at the end of each year, at the end of which year is account 1 worth more than account 2 for the first time?
Answer: C
The difference 500(1.06)ᵗ - (600 + 20t) is negative at t = 6 and positive at t = 7, so the graphs cross between 6 and 7 years (at about t ≈ 6.49). At the end of year 7, account 1 is worth more for the first time. Choice B picks the last year when account 1 is still behind ($709.26 < $720).
Question 14 of 20 · Multiple Choice
A graphing calculator is used to solve 1/x = x - 1. What are the approximate solutions?
Answer: B
The hyperbola y = 1/x meets the line y = x - 1 once in each branch. The exact values are (1 ± √5)/2, about 1.618 and -0.618. Check: 1/1.618 ≈ 0.618 = 1.618 - 1. Choice C has the signs reversed. Choice D includes x = 0, where 1/x is undefined.
Question 15 of 20 · Short Answer
Explain in your own words why the x-coordinates of the intersection points of y = f(x) and y = g(x) are the solutions of f(x) = g(x).
A point (a, b) on the graph of y = f(x) means f(a) = b. If the same point is on the graph of y = g(x), then g(a) = b too. So f(a) = g(a), which says x = a makes the equation true. Conversely, if f(a) = g(a), the point (a, f(a)) is on both graphs. The y-coordinate b is only the common value of the two sides.
Question 16 of 20 · Short Answer
Solve |2x + 3| = x + 6. Give the intersection points of the two graphs and check each solution.
x = 3 and x = -3, with intersections (3, 9) and (-3, 3). Case 1: 2x + 3 = x + 6 gives x = 3; check |9| = 9 = 3 + 6. Case 2: -(2x + 3) = x + 6 gives -3x = 9, so x = -3; check |-3| = 3 = -3 + 6.
Question 17 of 20 · Short Answer
Use successive approximations to find the positive solution of x³ = 2x + 2 to the nearest tenth. Show the values you tested.
x ≈ 1.8 x = 1.7: 1.7³ = 4.913 < 5.4. x = 1.8: 1.8³ = 5.832 > 5.6. So the solution is between 1.7 and 1.8. To round, test 1.75: 1.75³ ≈ 5.359 < 5.5, so the solution is between 1.75 and 1.8 and rounds to 1.8 (technology gives about 1.769).
Question 18 of 20 · Short Answer
Use technology to find all solutions of ln x = x - 1.5 to the nearest hundredth. Explain how you know you found all of them.
x ≈ 0.30 and x ≈ 2.36 Graph y = ln x and y = x - 1.5 and use the intersect feature twice. ln x is only defined for x > 0, and for large x the line grows much faster than ln x, so once the line passes the curve near x = 2.36 it stays above. Near x = 0, ln x drops toward negative infinity, below the line. There are exactly two intersections.
Question 19 of 20 · Short Answer
Solve 15/x = x + 2 by graphing y = 15/x and y = x + 2. Give the intersection points and verify the solutions.
x = 3 and x = -5, with intersections (3, 5) and (-5, -3). Check: 15/3 = 5 = 3 + 2 and 15/(-5) = -3 = -5 + 2. Multiplying by x (x ≠ 0) gives x² + 2x - 15 = 0, which factors as (x + 5)(x - 3) = 0.
Question 20 of 20 · Short Answer
Use a table to find the solution of 3ˣ = 8 - x to the nearest tenth.
x ≈ 1.7 x = 1: 3 < 7; x = 2: 9 > 6, so the solution is between 1 and 2. x = 1.6: 31.6 ≈ 5.80 < 6.4; x = 1.7: 31.7 ≈ 6.47 > 6.3, so the solution is between 1.6 and 1.7. x = 1.65: 31.65 ≈ 6.13 < 6.35, so it is between 1.65 and 1.7 and rounds to 1.7 (technology gives about 1.678).
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
Why are the x-coordinates the solutions and not the whole intersection point?
The equation f(x) = g(x) has only one variable, x, so its solutions are numbers, not points. At an intersection (a, b), both f(a) and g(a) equal b. The value a is what makes the equation true; b is just the common value of the two sides. Students who write the point instead of the x-value have usually mixed this standard up with solving a system in two variables.
Why use approximate methods at all? Can't we just solve algebraically?
Many equations in this standard cannot be solved exactly with high school algebra. For example, 2ˣ = x + 3 mixes an exponential and a linear function, and x³ = x + 1 is a cubic with no rational root. Graphs, tables and successive approximations give answers to any precision you need, and they also show how many solutions to look for.
Which function types does HSA.REI.D.11 require?
The standard lists linear, polynomial, rational, absolute value, exponential and logarithmic functions, for f, for g, or for both. A complete unit includes at least one equation of each type, such as |x - 2| = 0.5x + 1, x³ = x + 1, 6/x = x + 1, 2ˣ = x + 3 and log₂ x = x - 2.
How does a table of values tell me where a solution is?
Add a column for f(x) - g(x). If it is negative for one input and positive for the next, the graphs have crossed in between, as long as both functions are continuous on that interval. For 2ˣ and x + 3, the difference is -1 at x = 2 and 2 at x = 3, so a solution lies between 2 and 3.
When can a sign change be misleading?
When a function has a break, such as a rational function at a vertical asymptote. For 1/x = 0.5, the difference 1/x - 0.5 changes sign between x = -1 and x = 1 (it is -1.5 at x = -1 and 0.5 at x = 1), but the only solution is x = 2, outside that interval. The sign change happens because 1/x jumps at x = 0. Always look at a graph or check the answer.
How do I know I have found every solution with a graphing calculator?
Change the window. Zoom out to see where the graphs are headed, and think about end behavior: an exponential eventually outgrows any polynomial, and a logarithm is only defined for x > 0. The equation 2ˣ = x² is a classic trap: the standard window (-10 to 10 on both axes) shows x ≈ -0.77 and x = 2, but the third intersection, (4, 16), is above the top of the window, so students often miss x = 4.
What is a common mistake on this standard?
Reporting the y-coordinate, or the whole point, as the solution. Another common one is missing a solution because of the window or because only positive x-values were tested. With absolute value and rational equations, a third mistake is keeping an algebraic answer that does not check, such as x = -5 for |x| = 3x + 10.
How accurate do approximate answers need to be?
As accurate as the question asks, and students should say so, for example "x ≈ 2.44 to the nearest hundredth." To round correctly to the nearest hundredth by successive approximations, test the halfway value (such as 1.325) to decide which way to round.
Is HSA.REI.D.11 on the SAT?
Yes. The digital SAT includes a built-in graphing calculator, and questions in the Algebra and Advanced Math domains can be answered by graphing both sides of an equation and reading the x-coordinates of intersections. Students who understand why this works can use it as a check on algebraic work.
How does this connect to later courses?
The same idea is used to solve equations with trigonometric functions and, later, numerical methods in calculus that refine approximate solutions step by step. In the exponential and logarithmic unit, students also learn to solve some of these equations exactly, for example writing the solution of 5(2)ᵗ = 40 as a logarithm (HSF.LE.A.4).
07
Related Standards
6 standards
These standards connect to HSA.REI.D.11: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSA.REI.D.10Prerequisite
Understand that a graph of an equation is the set of all its solutions
