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1.11 Solving Equations and Inequalities Graphically
A solution to f(x)=g(x) is the x-coordinate of an intersection.
Worked example 1
01 / Use the two graphs to solve x2=2x+3. Report the intersection points and distinguish the requested solutions from their output values.pc1_11-G1-E1: y = x², y = 2x + 3
02 / Read both crossing points from the common axes.
03 / Their x-coordinates are −1 and 3, so these are the equation’s solutions.
04 / Check algebraically: x2−2x−3=(x+1)(x−3)=0.
05 / Substitution gives outputs 1 and 9, which are not the solution inputs.
Worked example 2 · Solve equations by intersections
The solid graph is y=∣x−1∣ and the dashed line is y=3. Solve their equality and explain why using only the right-hand crossing misses a solution.
pc1_11-G1-E2: y = |x − 1|, y = 3
The horizontal line meets both arms of the V-shaped graph. On the left, 1−x=3 gives x=−2; on the right, x−1=3 gives x=4. Both values have distance 3 from 1, so both satisfy the original equation.
Read the graphs
Solve equations by intersections
Use the two graphs to solve x2=2x+3. Report the intersection points and distinguish the requested solutions from their output values.pc1_11-G1-E1: y = x², y = 2x + 3
Read both crossing points from the common axes. Their x-coordinates are −1 and 3, so these are the equation’s solutions. Check algebraically: x2−2x−3=(x+1)(x−3)=0. Substitution gives outputs 1 and 9, which are not the solution inputs.
Answer: x=−1,3; intersections (−1,1) and (3,9).
Solve zero-form equations using x-intercepts
Solve x3−4x=0 using the x-intercepts, then verify that the graph shows all real roots.pc1_11-G2-E1: h(x) = x³ − 4x
Read the three x-intercepts at −2, 0, and 2. Factor x3−4x=x(x−2)(x+2). A real product is zero exactly when at least one factor is zero, proving the list complete.
Answer: x=−2,0,2.
Interpret graphical solutions in context
Two rental plans cost A(m)=18+4m and B(m)=42+2m dollars for m≥0 hours. Read the intersection and identify the cheaper plan before and after it.pc1_11-G3-E1: A: 18 + 4m, B: 42 + 2m
The graph intersection is (12,66), so equal cost occurs at 12 hours. The solid A line is below B for 0≤m<12 and above B for m>12. The difference A−B=2m−24 confirms the change in ordering.
Answer: Equal at 12 hours and 66 dollars; A is cheaper below 12, B above 12.
Solve inequalities from graphs
Read the graph to solve x2−x−6≤0. Explain both endpoint choices.pc1_11-G4-E1: h(x) = x² − x − 6
The curve crosses the axis at −2 and 3. It lies below the axis between these roots, so the interior belongs. The non-strict inequality also allows zero, so both roots are included.
Answer: [−2,3].
Approximate roots and assess viewing windows
The graph shows the positive root of x2=7. Use 2.645<7<2.646 and further checking if needed to report the root to two decimal places. Explain why 2.65 is not an exact equality.pc1_11-G5-E1: y = x², y = 7
The positive intersection lies between 2 and 3. The supplied bracket lies above the rounding boundary 2.645, so the nearest hundredth is 2.65. The exact root is 7, and 2.652=7.0225=7, so the rounded result must use an approximation sign.
Answer: x≈2.65.
Solve varied nonlinear graph families and respect their domains
Solve (x−1)4=16 algebraically and from the zero-form graph. Explain why there are two roots, then draw your own labeled graph showing the axis crossings.Use the labeled scale, curve styles, and domain information.
The difference graph meets the axis at −1 and 3. Taking the nonnegative fourth root gives ∣x−1∣=2, so x−1=2 or x−1=−2. Hence the two solutions are −1 and 3. The graph is symmetric about x=1, with minimum −16; the two branches rise to meet zero. A sketch must label both axes, these crossings, and the minimum.
Answer: x=−1,3.
15 questions covering every idea, with a worked answer for each.Start practice ↗
Textbook reference: Stewart, Redlin & Watson, Precalculus: Mathematics for Calculus, 7th ed. (Cengage, 2016). The explanations and practice here are original, organized by textbook section.