Graphing a quadratic inequality in two variables, a rope-strength model and systems of quadratic inequalities, solving a one-variable inequality by table and by graph, a robotics growth model, and the algebraic critical-value method.
Subject: Algebra 2 · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Algebra 2 · Chapter 4 — Quadratic Functions and Factoring
Graph and Solve Quadratic Inequalities
Objectives
Five outcomes. The first three are pictures; the last two are the algebra behind them.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.9 Graph and Solve Quadratic Inequalities §4.9, pp. 300-305 — the lesson these objectives are drawn from
Warm-up
Lesson 2.8 graphed linear inequalities in two variables. The procedure here is the same three steps.
Discussion prompt
Recall how you graphed y greater than 2x minus 1. What were the three steps, and which of them would change if the boundary were a parabola instead of a line?
Hint: Boundary, test point, shade.
Answer:
Graph the boundary, dashed for a strict inequality and solid otherwise; test a point not on the boundary; shade the side containing the point if it works, and the other side if it does not.
None of the three steps changes. Only the boundary is different — a curve rather than a line — and the two sides it separates are now the inside and the outside of a parabola rather than two half-planes.
Concept
A quadratic expression can only change from positive to negative by passing through zero. That single fact drives everything here: the parabola is the boundary, and on either side of it the inequality has one fixed verdict, which a single test point can determine.
quadratic inequality in two variables — An inequality that can be written as y compared with a x squared plus bx plus c, using less than, greater than, less than or equal to, or greater than or equal to. Its graph is all the ordered pairs that satisfy it.
\[ y > ax^2 + bx + c, \; y \ge ax^2 + bx + c, \; y < ax^2 + bx + c, \; y \le ax^2 + bx + c \]
The same reasoning in one variable puts the boundary on a number line instead of in the plane, and the roots of the corresponding equation become the endpoints of intervals.
Figure (svg): Two columns separating inequalities in two variables from inequalities in one variable
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.9 Graph and Solve Quadratic Inequalities §4.9, pp. 300-302
Section
Section 1
Concept
Graph the parabola, making it dashed for a strict inequality and solid when equality is allowed. Test any point not on the parabola: if it satisfies the inequality, shade the region containing it; if not, shade the other region.
\[ y > x^2 + 3x - 4 \]
The origin is the easiest test point whenever it is not on the boundary, because substituting zeros makes the arithmetic trivial.
Figure (svg): A dashed parabola with the region inside it shaded and a test point marked at the origin
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.9 Graph and Solve Quadratic Inequalities §4.9, pp. 300-300 — Graphing a Quadratic Inequality in Two Variables
Picture it
Example 1: y greater than x squared plus 3x minus 4.
Figure (svg): A dashed parabola with the region inside it shaded and a test point marked at the origin
The parabola is dashed because the symbol is strict, and the origin passes the test, so the region containing the origin — the inside — is shaded.
Worked example
Example 1, in the book's three steps.
\[ \text{Graph } y > x^2 + 3x - 4. \]
Graph the boundary parabola
Why: The related equation is y equals x squared plus 3x minus 4; its intercepts are at negative 4 and 1 and its vertex at negative 1.5.
Make it dashed
Why: The symbol is strictly greater than, so points on the parabola itself are not solutions.
Test a point inside
Why: At the origin the inequality reads 0 greater than negative 4, which is true.
\[ (0, 0)\text{ works} \]
Shade the region containing that point
Why: The origin lies inside the parabola, so the inside is shaded.
Figure (svg): The solution to Worked example graph a quadratic inequality shown as a ladder of expressions, one row per algebraic move
\[ y > x^2 + 3x - 4 \]
Verify: test a point on the other side
Why: At the point (0, negative 6) the inequality reads negative 6 greater than negative 4, which is false — so the outside really is excluded. Testing one point on each side is a complete check, because the verdict cannot change without crossing the boundary.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.9 Graph and Solve Quadratic Inequalities §4.9, pp. 300-300
Matching
The bar decides the style; the test point decides the side.
Match the pairs
Why: The four inequalities share a boundary, so the parabola is drawn identically each time; only its style and the shaded side differ. Note that inside and outside here mean above and below the curve, which for an upward parabola are the same thing.
Worked example
Guided Practice 1 to 3. Watch the direction of opening.
\[ \text{Graph } y > x^2 + 2x - 8, \; y \le 2x^2 - 3x + 1, \; y < -x^2 + 4x + 2. \]
First: dashed, opens up
Why: The symbol is strict, and testing the origin gives 0 greater than negative 8, which is true, so shade inside.
Second: solid, opens up
Why: The symbol allows equality. Testing the origin gives 0 less than or equal to 1, which is true, so shade the region containing the origin — the inside.
Third: solid or dashed?
Why: The symbol is strictly less than, so dashed; the parabola opens down because a is negative.
Test for the third
Why: At the origin, 0 less than 2 is true, and the origin lies inside a downward parabola whose vertex is at (2, 6), so shade inside.
Figure (svg): The solution to Worked example three more inequalities shown as a ladder of expressions, one row per algebraic move
\[ \text{three shaded regions} \]
Verify: check that inside is not automatic
Why: It happened three times here, but it is not a rule. For y less than x squared, the origin gives 0 less than 0, which is false, so the OUTSIDE is shaded. The test point decides every time; guessing from the symbol alone will eventually be wrong.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.9 Graph and Solve Quadratic Inequalities §4.9, pp. 301-301
Trap
\[ y \ge x^2 + 3x - 4 \]
Use a dashed curve because the inequality points one way
Why: The line style is chosen from whether the symbol is greater than or less than.
\[ \text{dashed parabola} \quad \text{(wrong)} \]
The points on the parabola itself do satisfy this inequality, since equality is allowed, so excluding them by drawing a dashed curve is incorrect.
\[ \ge \text{ and } \le \;\Longrightarrow\; \text{solid}; \qquad > \text{ and } < \;\Longrightarrow\; \text{dashed} \]
Choose the style from whether equality is included, not from the direction
Why: The direction decides which side to shade; the presence of the bar under the symbol decides the style.
\[ y \ge x^2 + 3x - 4 \;\Longrightarrow\; \text{solid parabola, shade inside} \]
Two independent decisions get made from two different features of the symbol, and the book flags this in an Avoid Errors note because they are easy to conflate.
Fill the middle
Example 1, at Step 2.
Fill in the blanks
y > x^2 + 3x - 4 \text-4 (0,0): \; 0 > 0 + 0 - 4, \text___ 0 > ___
Why: Substituting zero for x makes both variable terms vanish, leaving negative 4. Since 0 is greater than negative 4, the origin is a solution and the region containing it gets shaded. This is why the origin is the preferred test point whenever it is not on the boundary.
Ranking
Graphing a quadratic inequality in two variables.
Put in order
Why: The style is settled before any testing because it depends only on the symbol. The test point must be chosen off the curve, since a point on the boundary gives equality and cannot distinguish the two sides.
Prediction
Commit before reasoning.
Predict first
Would testing a point outside the parabola instead of the origin change the final graph?
Correct: No — a failing outside point and a passing inside point say the same thing.
\[ (0,-6): \; -6 > -4 \text{ is false} \;\Longrightarrow\; \text{shade the other side, the inside} \]
Why: The rule is to shade the region containing the test point when it works and the other region when it does not, so both choices lead to the same shading. Example 2's rope model uses a failing point and shades the opposite side, which is exactly this. The only bad choice is a point on the parabola itself, which gives equality and settles nothing.
Section
Section 2
Concept
A real model is graphed only over values that make sense, so a diameter or a time is drawn for non-negative inputs only. A system of quadratic inequalities is graphed by graphing each one and keeping the region common to all of them.
\[ W \le 1480d^2, \quad d \ge 0 \]
The overlap region of a system is called the graph of the system, exactly as for the linear systems of Lesson 3.3. A point belongs to it only if it satisfies every inequality.
Figure (svg): A solid parabola for the safe load of a rope, with the region beneath it shaded and a failing test point above it
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.9 Graph and Solve Quadratic Inequalities §4.9, pp. 301-301 — Examples 2 and 3
Picture it
Example 3: y at most negative x squared plus 4, and y greater than x squared minus 2x minus 3.
Figure (svg): Two overlapping shaded parabolic regions, with the region common to both marked
The solid downward parabola caps the region from above and the dashed upward one bounds it from below. Only the lens between them satisfies both.
Worked example
Example 2. A failing test point, and a restricted domain.
\[ \text{Graph } W \le 1480d^2 \text{ for a rope of diameter } d \text{ inches carrying } W \text{ pounds.} \]
Graph the boundary over a sensible domain
Why: The parabola W equals 1480 d squared is drawn only for non-negative d, since a diameter cannot be negative.
Choose the line style
Why: The symbol allows equality, so the parabola is solid: a load exactly at the limit is still safe.
Test a point inside the parabola
Why: At (1, 2000) the inequality reads 2000 at most 1480, which is false.
\[ (1, 2000)\text{ fails} \]
Shade the other region
Why: Since the inside fails, shade the region below the parabola.
Figure (svg): The solution to Worked example a rope's safe load shown as a ladder of expressions, one row per algebraic move
\[ W \le 1480d^2, \; d \ge 0 \]
Verify: test a point in the shaded region
Why: At (1, 1000) the inequality reads 1000 at most 1480, which is true, so a one-inch rope safely holds 1000 pounds. The graph is a design tool: for any diameter, the height of the curve is the largest safe load, and everything beneath it is permitted.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.9 Graph and Solve Quadratic Inequalities §4.9, pp. 301-301
Sorting
A point must satisfy both inequalities.
Sort into buckets
Test each point against y at most negative x squared plus 4 and y greater than x squared minus 2x minus 3.
The last of these is worth dwelling on: a point on a dashed boundary fails, which is precisely what dashed means.
Worked example
Example 3. Graph each, then intersect.
\[ \text{Graph the system } y \le -x^2 + 4 \text{ and } y > x^2 - 2x - 3. \]
Graph the first inequality
Why: The parabola opens down with vertex (0, 4), drawn solid, and the region on or inside it is shaded.
Graph the second
Why: The parabola opens up with intercepts at negative 1 and 3, drawn dashed, and the region strictly inside it is shaded.
Find where the two shadings overlap
Why: The overlap is the lens-shaped region between the two curves.
Describe the boundary of the overlap
Why: The upper edge is solid and included; the lower edge is dashed and excluded.
Figure (svg): The solution to Worked example a system of two inequalities shown as a ladder of expressions, one row per algebraic move
\[ \text{the overlap region} \]
Verify: test a point in the overlap
Why: At the origin the first reads 0 at most 4, true, and the second reads 0 greater than negative 3, also true, so the origin is in the graph of the system. Testing a point in only one region — say (0, 5), which fails the first — confirms that both conditions genuinely bite.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.9 Graph and Solve Quadratic Inequalities §4.9, pp. 301-301
Error analysis
A student graphs a system of two quadratic inequalities.
Annotate
On: \( \text{shade everything that satisfies } y \le -x^2 + 4 \text{ OR } y > x^2 - 2x - 3 \)
Same rule as the linear systems of Lesson 3.3: and means intersect. If a point fails even one inequality, it is out.
Fill the middle
Example 2, at Step 2.
Fill in the blanks
W \le 1480d^2 \text1480 (1, 2000): \; 2000 \le 1480(1)^2 = ___, \text___
Why: One squared is 1, so the right side is 1480, and 2000 is not at most 1480 — a one-inch rope cannot safely hold 2000 pounds. Because the test failed, the region NOT containing that point is shaded, which is the region below the curve.
Prediction
Commit before reasoning.
Predict first
The rope graph shows only non-negative diameters. Why?
Correct: A diameter cannot be negative, so that half has no meaning.
\[ W = 1480d^2 \text{ for } d \ge 0 \quad \text{(the model's domain)} \]
Why: The equation W equals 1480 d squared is perfectly well defined at negative d, and the parabola really does continue there — mathematically nothing is wrong with it. What restricts the graph is the situation being modelled, exactly as rejecting a negative time or length did in earlier lessons. Restricting the domain to the meaningful values is part of building an honest model.
Comparison
Fill the blanks. Lesson 3.3 against this lesson.
Comparison matrix
| Question | Linear system | Quadratic system |
|---|---|---|
| Boundaries | straight lines | parabolas |
| Regions each cuts out | two half-planes | inside and outside |
| The graph of the system | the overlap of all regions | the overlap of all regions |
| Shape of the solution | a polygon, possibly unbounded | a curved region such as a lens |
The method is unchanged from Lesson 3.3. Only the shape of the boundaries has changed, and with it the shape of the answer.
Section
Section 3
Concept
A quadratic inequality in one variable compares a quadratic with zero. Rewriting it so that zero is on one side turns the question into where the related parabola lies above or below the horizontal axis, which a table or a graph answers directly.
quadratic inequality in one variable — An inequality that can be written as a x squared plus bx plus c compared with zero. Its solution is a set of x values, usually one or two intervals.
\[ x^2 + x \le 6 \;\Longleftrightarrow\; x^2 + x - 6 \le 0 \]
A table finds the endpoints exactly only when they happen to be integers. A graph handles any endpoints, and the quadratic formula gives them exactly.
Figure (svg): A table of values for x squared plus x minus six, with the non-positive entries highlighted
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.9 Graph and Solve Quadratic Inequalities §4.9, pp. 302-302 — Solve a quadratic inequality using a table
Picture it
Example 4: x squared plus x at most 6.
Figure (svg): A table of values for x squared plus x minus six, with the non-positive entries highlighted
The expression is at or below zero from negative 3 to 2 inclusive, and the two endpoints are exactly where the value is zero — which is why the roots of the equation are always the boundary.
Worked example
Example 4. Rewrite first, then tabulate.
\[ \text{Solve } x^2 + x \le 6 \text{ using a table.} \]
Rewrite with zero on one side
Why: Subtracting 6 gives the expression whose sign is being asked about.
\[ x ^{2} + x - 6 \le 0 \]
Tabulate integer values
Why: From negative 5 to 4 the values are 14, 6, 0, negative 4, negative 6, negative 6, negative 4, 0, 6, 14.
Read where the values are at or below zero
Why: That happens from x equal to negative 3 through x equal to 2.
\[ -3\text{ to } 2 \]
Decide about the endpoints
Why: At negative 3 and at 2 the value is exactly zero, which satisfies at most zero, so both are included.
Figure (svg): The solution to Worked example solve with a table shown as a ladder of expressions, one row per algebraic move
\[ -3 \le x \le 2 \]
Verify: test a value inside and one outside
Why: At x equal to 0 the original reads 0 at most 6, true. At x equal to 3 it reads 12 at most 6, false. The table's symmetry about negative 0.5 is also visible, and that value is negative b over 2a — the vertex, as it should be.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.9 Graph and Solve Quadratic Inequalities §4.9, pp. 302-302
Fill the middle
Example 4, at the first step.
Fill in the blanks
x^2 + x \le 6 \;\Longleftrightarrow\; x^2 + x - 6 \le 0
Why: Subtracting 6 from both sides leaves zero on the right, which is the form the table and graph methods both want. Note that subtracting the same number from both sides never flips an inequality; only multiplying or dividing by a negative does that.
Worked example
Example 5. The endpoints are irrational, so a table cannot find them.
\[ \text{Solve } 2x^2 + x - 4 \ge 0 \text{ by graphing.} \]
State what the graph must show
Why: The solution is the set of x for which the graph of y equals 2x squared plus x minus 4 lies on or above the horizontal axis.
Find the x-intercepts with the formula
Why: Setting y to zero gives negative 1 plus or minus the root of 33, over 4.
\[ x = \frac{-1 + - \sqrt{33}}{4} \]
Evaluate them
Why: The root of 33 is about 5.745, giving about 1.19 and about negative 1.69.
\[ -1.69\text{ and } 1.19 \]
Read the intervals
Why: The parabola opens up, so it is on or above the axis outside the intercepts.
\[ x \le - 1.69\text{ or } x \ge 1.19 \]
Figure (svg): The solution to Worked example solve by graphing shown as a ladder of expressions, one row per algebraic move
\[ x \le \tfrac{-1-\sqrt{33}}{4} \quad \text{or} \quad x \ge \tfrac{-1+\sqrt{33}}{4} \]
Verify: test one value in each of the three intervals
Why: At x equal to negative 2: 8 minus 2 minus 4 is 2, which is at least zero, so the left interval is in. At x equal to 0: negative 4, which is not, so the middle is out. At x equal to 2: 8 plus 2 minus 4 is 6, in. Three tests, three intervals, and the verdicts match the graph.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.9 Graph and Solve Quadratic Inequalities §4.9, pp. 302-302
Error analysis
A student solves an inequality using a table of the wrong expression.
Annotate
On: \( x^2 + x \le 6: \text{ tabulate } x^2 + x \text{ and look for values at or below } 0 \)
Decide which form you are working in and stay in it. Rewriting so that zero is on one side is the safer habit, because it is what the graph method needs anyway.
Sorting
An upward parabola is negative between its intercepts and positive outside them.
Sort into buckets
Sort each inequality by where its solution lies, given that each opens upward.
For a downward parabola every verdict here reverses, which is why reading the sign of a is the first thing to do.
Prediction
Commit before reasoning.
Predict first
Example 5's endpoints are about negative 1.69 and 1.19. What would a table of integers have shown?
Correct: The right intervals but with wrong, integer endpoints.
\[ x = \tfrac{-1 \pm \sqrt{33}}{4}: \; \text{no table of integers will produce this} \]
Why: A table of integers would show the expression positive at negative 2, negative at negative 1, 0 and 1, and positive at 2 — correctly locating the two sign changes but placing the endpoints only within a unit. That is enough for a rough answer and useless for an exact one. The graph plus the quadratic formula gives the endpoints exactly, which is why the book uses a table only when the roots turn out to be integers.
Comparison
Fill the blanks. Two ways to answer the same question.
Comparison matrix
| Question | Table | Graph |
|---|---|---|
| What you build | a list of values | a sketch of the parabola |
| How endpoints are found | by spotting where the value is 0 | as the x-intercepts |
| Works for irrational endpoints | no | yes, with the formula |
| Effort | ten evaluations | one use of the quadratic formula |
The table is a good way to see what an inequality means; the graph is the practical method once you trust it.
Section
Section 4
Concept
When a model asks when a quantity exceeded some level rather than when it equalled it, the result is an inequality. Rewrite it with zero on one side, find the boundary values, and report the interval — intersected with whatever domain the model is valid on.
\[ T(x) > 100 \;\Longrightarrow\; 7.51x^2 - 16.4x - 65 > 0 \]
The model's stated domain matters as much as the algebra. A boundary value outside that domain does not become part of the answer.
Figure (svg): A parabola with two irrational x-intercepts, showing the two rays on which it lies on or above the axis
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.9 Graph and Solve Quadratic Inequalities §4.9, pp. 303-303 — Use a quadratic inequality as a model
Picture it
Example 5's parabola, showing the shape of every answer of this kind.
Figure (svg): A parabola with two irrational x-intercepts, showing the two rays on which it lies on or above the axis
Once the boundary values are known, the answer is read off in one glance: outside them for an upward parabola asked to be non-negative, between them otherwise.
Worked example
Example 6. The domain is part of the answer.
\[ \text{With } T(x) = 7.51x^2 - 16.4x + 35.0 \text{ for } 0 \le x \le 9, \text{ find when } T(x) > 100. \]
Write the inequality
Why: The number of teams is to exceed 100.
\[ 7.51 x ^{2} - 16.4 x + 35.0 > 100 \]
Move everything to one side
Why: Subtracting 100 gives a constant of negative 65.
\[ 7.51 x ^{2} - 16.4 x - 65 > 0 \]
Find the boundary on the model's domain
Why: Graphing on 0 to 9 shows one x-intercept, at about 4.2.
\[ x\text{ is about } 4.2 \]
Read the interval and intersect with the domain
Why: The graph is above the axis for x greater than 4.2, and the model runs only to x equal to 9.
\[ 4.2 < x \le 9 \]
Translate back into years
Why: X is years since 1992, so the whole years in that range are 5 through 9.
\[ 1997\text{ to } 2001 \]
Figure (svg): The solution to Worked example when were there more than 100 teams shown as a ladder of expressions, one row per algebraic move
\[ 4.2 < x \le 9 \;\Longrightarrow\; 1997\text{--}2001 \]
Verify: evaluate the model at a year in and a year out
Why: At x equal to 4, the model gives 7.51 times 16, minus 65.6, plus 35, which is about 89 teams — under 100, correctly outside the interval. At x equal to 5 it gives about 140 teams, correctly inside. The other root of the quadratic is negative and lies outside the model's domain entirely, which is why only one intercept appeared.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.9 Graph and Solve Quadratic Inequalities §4.9, pp. 303-303
Ranking
Answering a when-was-it-more-than question.
Put in order
Why: The last step does two jobs and both are easy to skip: cutting the interval down to the domain, and converting an x value back into a year, a price or a length. An answer left as 4.2 less than x is not yet an answer to the question that was asked.
Worked example
Guided Practice 6. The same model, a higher threshold.
\[ \text{Find when } T(x) \ge 200. \]
Write and rearrange
Why: Subtracting 200 gives a constant of negative 165.
\[ 7.51 x ^{2} - 16.4 x - 165 \ge 0 \]
Apply the quadratic formula
Why: The radicand is 16.4 squared plus 4 times 7.51 times 165, which is about 5225.6.
\[ \sqrt{5225.6}\text{ is about } 72.3 \]
Evaluate the positive root
Why: 16.4 plus 72.3, over 15.02, is about 5.9; the other root is negative and outside the domain.
\[ x\text{ is about } 5.9 \]
Read the interval and translate
Why: The whole years from 6 to 9 satisfy it, which are 1998 through 2001.
\[ 1998\text{ to } 2001 \]
Figure (svg): The solution to Worked example at least 200 teams shown as a ladder of expressions, one row per algebraic move
\[ 5.9 \le x \le 9 \;\Longrightarrow\; 1998\text{--}2001 \]
Verify: check the boundary year
Why: At x equal to 5 the model gives about 140 teams, below 200, and at x equal to 6 about 207, above it. So the crossing really is between those years, consistent with the root at about 5.9. Raising the threshold from 100 to 200 moved the start by only one year, because the quadratic is growing quickly by then.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.9 Graph and Solve Quadratic Inequalities §4.9, pp. 303-303
Error analysis
A student solves the robotics inequality and reports both intervals.
Annotate
On: \( 7.51x^2 - 16.4x - 65 > 0 \;\Longrightarrow\; x < -2.0 \text{ or } x > 4.2 \)
A model's domain is part of the model. Solve the inequality first, then intersect with the domain, and never report an interval the model does not cover.
Fill the middle
Example 6, at the second step.
Fill in the blanks
7.51x^2 - 16.4x + 35.0 > 100 \;\Longrightarrow\; 7.51x^2 - 16.4x - 65 > 0
Why: Subtracting 100 from both sides turns the constant 35 into negative 65. Only the constant changes, since the other terms contain x, and the direction of the inequality is unaffected because subtraction never flips it.
Prediction
Commit before reasoning.
Predict first
A quadratic inequality usually has two boundary values. Why does Example 6 report only one?
Correct: The other boundary is negative and lies outside the model's domain.
\[ x < -2.0 \text{ or } x > 4.2, \; \text{intersected with } 0 \le x \le 9 \]
Why: The quadratic 7.51x squared minus 16.4x minus 65 does have two roots, at about negative 2.0 and about 4.2, so the inequality holds on two intervals. But the model is only defined from 0 to 9, and the interval to the left of negative 2 lies entirely outside that. Intersecting with the domain removes it, leaving one boundary visible.
Two truths and a lie
All three are about modelling with inequalities.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. A quadratic inequality generally yields two intervals, and only the model's domain reduced this one to a single interval. Example 7's answer, x less than negative 3 or x greater than 5, is two intervals with no domain restriction to trim it.
Section
Section 5
Concept
Replace the inequality sign with an equals sign and solve. Those roots are the critical values: plot them on a number line, where they cut it into intervals. Test one value from each interval in the original inequality, and keep the intervals that pass.
critical values — The solutions of the equation obtained by replacing the inequality sign with an equals sign. They are the only places where the expression can change sign, so they are the endpoints of the solution intervals.
\[ x^2 - 2x > 15 \;\Longrightarrow\; (x+3)(x-5) = 0 \;\Longrightarrow\; x = -3, \; 5 \]
Use open dots when the inequality is strict and closed dots when equality is allowed, exactly as on the number lines of Lesson 1.6.
Figure (svg): A number line with two critical values dividing it into three intervals, each with a test value and its verdict
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.9 Graph and Solve Quadratic Inequalities §4.9, pp. 303-303 — Solve a quadratic inequality algebraically
Picture it
Example 7: x squared minus 2x greater than 15.
Figure (svg): A number line with two critical values dividing it into three intervals, each with a test value and its verdict
The outer intervals pass and the middle one fails, so the solution is x less than negative 3 or x greater than 5. Two critical values always give three intervals.
Worked example
Example 7. No graph is drawn at all.
\[ \text{Solve } x^2 - 2x > 15 \text{ algebraically.} \]
Replace the inequality with an equation
Why: Solving x squared minus 2x equals 15 will locate the sign changes.
\[ x ^{2} - 2 x - 15 = 0 \]
Solve it
Why: The trinomial factors as x plus 3 times x minus 5.
\[ x = -3\text{ or } x = 5 \]
Plot the critical values with open dots
Why: The symbol is strict, so neither value is itself a solution, and the line is cut into three intervals.
Test one value in each interval
Why: At negative 4 the expression is 24, greater than 15; at 1 it is negative 1, which is not; at 6 it is 24, which is.
Write the solution
Why: The two outer intervals pass.
\[ x < -3\text{ or } x > 5 \]
Figure (svg): The solution to Worked example solve algebraically shown as a ladder of expressions, one row per algebraic move
\[ x < -3 \quad \text{or} \quad x > 5 \]
Verify: check a critical value itself
Why: At x equal to 5 the expression is 25 minus 10, or 15, which is not strictly greater than 15 — so 5 is correctly excluded and the open dot is right. Had the inequality been at least 15, both critical values would join the solution and the dots would be closed.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.9 Graph and Solve Quadratic Inequalities §4.9, pp. 303-303
Fill the middle
Example 7, at the equation.
Fill in the blanks
x^2 - 2x - 15 = 0 \;\Longrightarrow\; (x + 3)(x - 5) = 0
Why: Three times negative 5 is negative 15 and 3 plus negative 5 is negative 2, so the second bracket is x minus 5. The critical values are the opposites of the constants in the brackets, negative 3 and 5, exactly as for the roots in Lesson 4.3.
Worked example
Guided Practice 7. The critical values are not both integers.
\[ \text{Solve } 2x^2 - 7x > 4 \text{ algebraically.} \]
Form and solve the equation
Why: Subtracting 4 gives 2x squared minus 7x minus 4 equals zero.
\[ 2 x ^{2} - 7 x - 4 = 0 \]
Factor it
Why: Two is prime and negative 4 gives opposite signs; the pair 1 and negative 4 placed as 2x plus 1 and x minus 4 gives a middle term of negative 7x.
\[ (2 x + 1) (x - 4) = 0 \]
Read the critical values
Why: The brackets vanish at negative one half and at 4.
\[ x = -\frac{1}{2}\text{ or } x = 4 \]
Test the three intervals
Why: At negative 1: 2 plus 7 is 9, greater than 4, pass. At 0: 0, not greater than 4, fail. At 5: 50 minus 35 is 15, pass.
Write the solution with open dots
Why: The symbol is strict, so both critical values are excluded.
\[ x < -\frac{1}{2}\text{ or } x > 4 \]
Figure (svg): The solution to Worked example a leading coefficient shown as a ladder of expressions, one row per algebraic move
\[ x < -\tfrac{1}{2} \quad \text{or} \quad x > 4 \]
Verify: sanity-check against the shape
Why: The parabola opens upward, so it is positive outside its intercepts — which matches the two outer intervals passing. Whenever the algebraic method gives a middle interval passing for an upward parabola, something has gone wrong, so this shape check is worth a second.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.9 Graph and Solve Quadratic Inequalities §4.9, pp. 303-303
Trap
\[ x^2 - 2x > 15 \]
Divide both sides by x to simplify
Why: The x is cancelled from the two left-hand terms.
\[ x - 2 > \tfrac{15}{x} \quad \text{(wrong and worse)} \]
Dividing by x is illegal when x can be negative, because that reverses the inequality, and impossible when x is zero. The result is not simpler either.
\[ x^2 - 2x - 15 > 0 \;\Longrightarrow\; (x+3)(x-5) > 0 \]
Collect on one side and find the critical values instead
Why: The sign of a product is determined by testing intervals, not by dividing.
\[ x < -3 \quad \text{or} \quad x > 5 \]
The same warning applied to equations in Lesson 4.4, where dividing by a variable silently lost a root. In an inequality it can also flip the direction, so it goes wrong two ways at once.
Sorting
Ask whether the critical value itself satisfies the inequality.
Sort into buckets
Sort each inequality by how its critical values are marked.
The dot style follows the same rule as the dashed or solid parabola in the two-variable case, and for the same reason.
Prediction
Commit before reasoning.
Predict first
The algebraic method tests just one value in each interval. Why is that sufficient?
Correct: Between critical values the expression cannot change sign.
\[ \text{sign change} \;\Longrightarrow\; \text{a zero in between} \]
Why: A quadratic is continuous, so to move from positive to negative it must pass through zero — and the critical values are precisely where it is zero. Within an interval containing no zeros, the sign is therefore constant, and a single test reveals it. This is why the method is exact rather than a sampling estimate, and the same argument extends to the higher-degree inequalities of Chapter 5.
Comparison
Fill the blanks. All three give the same answer.
Comparison matrix
| Method | What you produce | Best when |
|---|---|---|
| Table | a list of values | the endpoints are integers |
| Graph | a sketch with the x-intercepts marked | you want to see the shape |
| Algebraic | critical values and interval tests | you want exact endpoints quickly |
| All three | the same intervals | they rest on the same fact |
The algebraic method is usually fastest, but sketching the parabola alongside it is a five-second check that the right intervals were kept.
Comparison
Fill the blanks. Same idea, two settings.
Comparison matrix
| Question | Two variables | One variable |
|---|---|---|
| The boundary is | a parabola in the plane | two points on a number line |
| The answer is | a shaded region | one or two intervals |
| Included boundary shown by | a solid curve | closed dots |
| How you decide | test a point off the curve | test a value in each interval |
Both rest on the same fact: an expression cannot change sign without passing through zero, so the zeros are the only possible boundaries.
Pattern
Two routines, depending on how many variables there are.
A shape check costs nothing: an upward parabola is negative between its intercepts and positive outside them, and a downward one is the reverse.
OpenStax Algebra and Trigonometry 2e, §5.3 Graphs of Polynomial Functions §5.3
Check
Graphing in two variables. Two decisions to make.
Check your understanding
How should y >= x^2 + 3x - 4 be graphed?
Answer: A
Why: The symbol allows equality, so the curve is solid; testing (0, 0) gives 0 >= -4, which is true, so the region containing the origin is shaded.
Check
One variable, by table or graph.
Check your understanding
Solve x^2 + x <= 6.
Answer: A
Why: Rewriting gives x^2 + x - 6 <= 0, whose critical values are -3 and 2; an upward parabola is at or below zero between them.
Check
The algebraic method.
Check your understanding
Solve x^2 - 2x > 15.
Answer: A
Why: The critical values are -3 and 5; testing -4, 1 and 6 shows the two outer intervals pass.
Real world
A company's daily profit, in dollars, from selling x units is modelled by P of x equals negative 2x squared plus 120x minus 1000, valid for 0 at most x at most 60.
Discussion prompt
For which numbers of units is the company profitable, that is P of x greater than 0? And for which is the profit at least 500 dollars?
Hint: Divide by negative 2 carefully, or use the quadratic formula directly.
Answer:
\[ -2x^2 + 120x - 1000 > 0 \;\Longleftrightarrow\; x^2 - 60x + 500 < 0 \]
\[ x^2 - 60x + 500 = 0 \;\Longrightarrow\; x = 10 \text{ or } 50 \;\Longrightarrow\; 10 < x < 50 \]
\[ P(x) \ge 500 \;\Longrightarrow\; x^2 - 60x + 750 \le 0 \;\Longrightarrow\; x = 30 \pm 5\sqrt{6} \approx 17.8, 42.2 \]
The company is profitable for between 10 and 50 units, and makes at least 500 dollars for about 18 to 42 units.
Two things are worth noticing. Dividing by negative 2 flipped the inequality, which is the one step in this lesson where the direction changes. And the answer is an interval rather than two rays because the parabola opens downward — selling too many units is as unprofitable as selling too few, which the model captures and a linear one could not.
Commit first
Answer, then rate your confidence honestly.
Predict first
You solve a quadratic inequality and get the middle interval as your answer, for a parabola that opens upward and asks for greater than zero. What has happened?
Correct: An error — an upward parabola is negative between its intercepts.
\[ a > 0: \; \text{negative between the roots, positive outside} \]
Why: For an upward parabola the graph dips below the axis exactly between the two x-intercepts, so a greater-than-zero question must be answered by the two outer intervals. Getting the middle one means either the tests were done in the wrong inequality or their verdicts were recorded backwards. This shape check takes a second and catches the most common error in the algebraic method, which is why it is worth doing every time.
Explain it
They can solve quadratic equations but have never seen a quadratic inequality.
Discussion prompt
In four sentences or fewer, explain why solving the equation is the first step in solving the inequality, and what you do after that.
Hint: Talk about where a sign can change.
Answer:
A quadratic expression is a smooth curve, so it can only go from positive to negative by passing through zero. Solving the equation finds every place where it is zero, and those are the only places its sign can change.
Those values cut the number line into intervals, and inside each interval the sign is fixed. So you test one number from each interval, and keep the whole interval if the test passes.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For dashed and solid, read the bar under the symbol; for the side, always test a point rather than reasoning. For intervals, sketch the parabola and check that your answer matches its shape. For domains, write the model's stated range at the top of the page before starting. For dots, ask whether the critical value itself satisfies the inequality — substitute it and see. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Take the single quadratic x squared minus 2x minus 15 and answer four questions about it on one page. Top left: graph y greater than x squared minus 2x minus 15 in the plane, with the right line style, a labelled test point and the correct region shaded. Top right: solve x squared minus 2x minus 15 greater than 0 algebraically, drawing the number line with critical values, dot styles and the three test values with their verdicts. Bottom left: sketch the parabola alone and shade in one colour where it is above the axis and in another where it is below, then write the two inequalities those regions answer. Bottom right: change the inequality to at least zero and write what changes in each of your three panels and what does not. In a margin, write the one sentence about sign changes that justifies testing only one value per interval.
If your top-right and bottom-left panels disagree about which intervals are positive, trust the sketch and recheck your test values — the shape of an upward parabola is not negotiable.
Recap
Five things, all consequences of one fact about sign changes.
| If you see | Then |
|---|---|
| A strict symbol | Dashed curve, or open dots |
| A symbol with a bar | Solid curve, or closed dots |
| A system | Keep only the overlap |
| An upward parabola, asked for < 0 | The answer is between the intercepts |
| An upward parabola, asked for > 0 | The answer is the two outer intervals |
| A stated domain | Intersect your interval with it |
Lesson 4.10 closes the chapter by running everything backwards: given points a parabola passes through, find the quadratic function itself.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.9 Graph and Solve Quadratic Inequalities §4.9, pp. 300-305 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Algebra 2 — $55/session, free consultation.