Sketching the graph of a quadratic inequality in two variables. Includes the four kinds of quadratic inequality, deciding whether a point lies inside or outside a parabola, using a dashed or solid boundary curve, the test-point method for choosing which region to shade, and the above-or-below method that reads the region straight off the inequality symbol.
Subject: Algebra 1 · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Algebra 1 · Chapter 9 — Quadratic Equations and Functions
Graphing Quadratic Inequalities
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.8 Graphing Quadratic Inequalities §9.8, pp. 547-552 — the lesson these objectives are drawn from
Warm-up
Lesson 6.8 graphed linear inequalities in two variables. Everything about the method carries over; only the boundary changes shape.
Discussion prompt
Describe how you graphed y less than two x plus one. Which parts of that method would still work if the boundary were a parabola?
Hint: Boundary, then a decision about which side.
Answer:
You drew the boundary line, made it dashed because the symbol was strict, tested a point not on it, and shaded the side that worked.
Every one of those steps survives. The boundary becomes a curve rather than a line, the plane is still divided into two regions, and a test point still decides between them — so this lesson is Lesson 6.8 with a different boundary drawn.
Concept
The graph of a quadratic inequality consists of all the ordered pairs that are solutions of the inequality. That is a whole region of the plane rather than a set of isolated points.
graph of a quadratic inequality — The set of all ordered pairs that satisfy the inequality, shown as a shaded region of the coordinate plane bounded by the parabola of the corresponding equation.
There are four kinds, one for each inequality symbol.
Figure (svg): The four kinds of quadratic inequality
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.8 Graphing Quadratic Inequalities §9.8, pp. 547-547
Section
Section 1
Concept
A parabola divides the plane into two regions, one inside the curve and one outside. To decide which contains a point, compare the point's y-coordinate with the value of the function at the same x.
No accurate drawing is needed, only one substitution.
Figure (svg): Three points tested against a parabola
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.8 Graphing Quadratic Inequalities §9.8, pp. 547-547 — Example 1, Check Points
Picture it
Two inside, one outside.
Figure (svg): Three points tested against a parabola
The point C sits below the curve at its own position, which for an upward parabola puts it outside. The comparison is local: what matters is the curve's height directly beneath or above the point, not anywhere else.
Worked example
This is Example 1 from the textbook.
\[ \text{For } y = x^2 - 3x - 3, \text{ are } A(3, -2), \; B(-1, 4) \text{ and } C(4, -3) \text{ inside or outside?} \]
Test A
Why: At x equal to three the curve is at negative three.
\[ -2 > -3, \text{ inside} \]
Test B
Why: At x equal to negative one the curve is at one.
\[ 4 > 1, \text{ inside} \]
Test C
Why: At x equal to four the curve is at one.
\[ -3 < 1, \text{ outside} \]
State the pattern
Why: Above the curve is inside, here.
Figure (svg): Three points tested against a parabola
\[ A, \; B \text{ inside}; \quad C \text{ outside} \]
Verify: check A against the vertex
Why: The vertex is at x equal to 1.5, where the value is negative 5.25, so the whole inside region lies above that level near the bottom. A at negative two is comfortably above the curve at its own x, which is what inside means.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.8 Graphing Quadratic Inequalities §9.8, pp. 547-547
Sorting
For the upward parabola y equals x squared minus four x plus three.
Sort into buckets
Sort each point by where it lies.
Two points share the same x here and land in different regions, which is a reminder that the comparison is with the curve's height at that x rather than with any fixed level.
Worked example
Guided Practice 1 to 3, on a different parabola.
\[ \text{For } y = x^2 - 4x + 3, \text{ classify } A(1, 2), \; B(0, 0) \text{ and } C(2, 1). \]
Test A
Why: At x equal to one the curve is at nought.
\[ 2 > 0, \text{ inside} \]
Test B
Why: At x equal to nought the curve is at three.
\[ 0 < 3, \text{ outside} \]
Test C
Why: At x equal to two the curve is at negative one.
\[ 1 > -1, \text{ inside} \]
Note the vertex
Why: The lowest point is at two, negative one.
\[ C \text{ sits above it} \]
Figure (svg): Three points tested against a parabola
\[ A, \; C \text{ inside}; \quad B \text{ outside} \]
Verify: check B, which is at the origin
Why: At x equal to nought the curve is at three, well above the origin, so the origin lies below the curve and therefore outside. The origin is often the easiest point to test, and here it lands outside.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.8 Graphing Quadratic Inequalities §9.8, pp. 547-547
Trap
The point (4, -3) looks close to the curve, so it is hard to say which side it is on.
Decide by looking at the drawing
Why: The sketch is what the question supplied, so the sketch is what got read.
A sketch cannot resolve a point close to the boundary. Substituting settles it exactly: the curve is at one when x is four, and negative three is below that, so the point is outside with no doubt at all.
\[ f(4) = 16 - 12 - 3 = 1 \quad \text{and} \quad -3 < 1 \]
Substitute the point's x and compare the two heights
Why: Arithmetic beats eyesight near a boundary.
The drawing is worth having for the overall picture and never for a close call.
Faded example
Substitute the x, then compare.
Fill in the blanks
y = x^2 - 3x - 3 \text1 x = 4: \; f(4) = below; \quad \text___ (4, -3) \text___ ___ \text___
Why: For an upward parabola, below the curve means outside it. The words above and below are unambiguous, while inside and outside swap meaning when the parabola opens the other way.
Prediction
Inside and outside for a downward curve.
Predict first
For a parabola opening down, which points are inside?
Correct: Those below the curve.
Method II shades above or below and never mentions inside or outside at all.
Why: Inside means within the bowl, and a downward parabola's bowl opens downwards, so its interior is underneath. That is why above and below are safer words than inside and outside: they mean the same thing whichever way the curve opens, and the shading rules in this lesson are stated in those terms for exactly that reason.
Socratic
The regions are large and the point is one place.
Discussion prompt
Explain why comparing one pair of heights decides which of two regions contains a point. Then say what would go wrong if the two heights were equal.
Hint: Think about the vertical line through the point.
Answer:
Every point of the plane sits on a vertical line, and that line meets the parabola exactly once. So there are only two possibilities for a point on it: above that meeting place or below it, and those correspond precisely to the two regions the curve divides the plane into.
If the two heights were equal the point would be on the parabola itself, in neither region. That is not a failure of the method but a third case, and it is exactly the case that the dashed-or-solid decision in the next section is about — whether such points count as solutions.
Section
Section 2
Concept
Sketch a dashed parabola for inequalities with a strict symbol, to show that points on the curve are not solutions. Sketch a solid parabola for an or-equal symbol, to show that they are.
The same convention as for linear inequalities.
Figure (svg): A dashed boundary against a solid one
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.8 Graphing Quadratic Inequalities §9.8, pp. 548-548 — Method I, step on dashed and solid parabolas
Picture it
A strict symbol excludes the curve.
Figure (svg): A dashed boundary against a solid one
The two curves are in exactly the same place; only their style differs. That style is carrying real information about which ordered pairs are solutions.
Worked example
Four inequalities, four decisions.
\[ \text{Which boundary style for } y < x^2, \; y \le x^2, \; y > x^2 \text{ and } y \ge x^2? \]
Take the first
Why: A strict less than.
Take the second
Why: Less than or equal to.
Take the third
Why: A strict greater than.
Take the fourth
Why: Greater than or equal to.
Figure (svg): A dashed boundary against a solid one
\[ \text{dashed}, \; \text{solid}, \; \text{dashed}, \; \text{solid} \]
Verify: test a point on the curve
Why: The point (2, 4) lies on y equals x squared. It satisfies y less than or equal to x squared, since four equals four, but not y less than x squared, since four is not less than four. That is precisely the difference the two styles record.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.8 Graphing Quadratic Inequalities §9.8, pp. 548-548
Matching
Strict or or-equal.
Match the pairs
Why: Only whether the symbol carries an equals part matters for the style; the direction is irrelevant to it. The direction decides the shading, which is a separate question answered separately.
Worked example
One point that always sits on the boundary.
\[ \text{Is the vertex of } y = x^2 \text{ a solution of } y \ge x^2? \text{ And of } y > x^2? \]
Find the vertex
Why: The lowest point of the curve.
\[ (0, 0) \]
Test the first
Why: Is nought at least nought?
Test the second
Why: Is nought greater than nought?
Draw the conclusion
Why: One includes it, the other does not.
Figure (svg): A dashed boundary against a solid one
\[ 0 \ge 0 \;\checkmark \qquad 0 > 0 \;\times \]
Verify: say why any point on the curve behaves the same way
Why: Every point on the curve has its y exactly equal to the function's value, so an or-equal symbol is satisfied and a strict one is not. The vertex is nothing special here — it just happens to be the easiest such point to name.
Trap
The symbol is less than, so the curve is dashed and the shading goes to the left.
Read less than as pointing left, as on a number line
Why: Less than meant left in Chapter 6, so it was carried over.
For a two-variable inequality with y isolated, less than means below rather than left. The number line's left-and-right meaning belongs to inequalities in one variable.
The symbol is less than, so the curve is dashed and the shading goes below.
Read the symbol as a comparison of y with the expression
Why: Below means smaller y, which is what less than asks for.
The dashed decision and the shading decision come from the same symbol but mean different things.
Faded example
Style, then side.
Fill in the blanks
For y at most the expression, draw a solid curve and shade below it.
Why: The equals part gives the style and the direction gives the side, and the two are read off independently. Confusing them produces a graph that is right in one respect and wrong in the other.
Elimination
Of the inequality y greater than x squared.
Eliminate the wrong options
Which ordered pair satisfies it?
Survives elimination: A
Why: Only the first has its y strictly above the curve's value of four. Two of the wrong answers are on the boundary itself, which a strict symbol excludes, and that is exactly what a dashed curve is drawn to record.
Socratic
The shading is the main thing.
Discussion prompt
Explain what information the dashed style carries that the shading alone cannot. Then say when the distinction genuinely changes an answer.
Hint: Think about points exactly on the curve.
Answer:
The shading says which region's points are solutions, but points on the curve itself are in neither region. The style is the only place the graph records whether those boundary points count, so without it the picture would be ambiguous about an entire curve's worth of ordered pairs.
It matters whenever a question asks about a specific point that happens to lie on the boundary, or when a situation has a limit that can be met exactly — a budget that may be spent exactly, a height that may be reached exactly. In those cases the difference between at most and less than is the difference between the boundary case being allowed and forbidden, and that is usually the case people care about most.
Section
Section 3
Concept
The parabola separates the plane into two regions. Test any point not on the curve: if it satisfies the inequality, shade its region, and if not, shade the other one.
The origin is a good test point when it is not on the curve.
Figure (svg): Testing a point to decide which region to shade
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.8 Graphing Quadratic Inequalities §9.8, pp. 548-548 — Method I: Graphing a Quadratic Inequality, Example 2, and its Study Tip on the origin
Picture it
The point (1, 2) does not work.
Figure (svg): Testing a point to decide which region to shade
A failing test is just as informative as a passing one. It rules out one region and therefore selects the other, so the substitution is never wasted.
Worked example
This is Example 2 from the textbook.
\[ \text{Sketch the graph of } y < 2x^2 - 3x. \]
Sketch the boundary
Why: Opens up, vertex at three quarters.
\[ \left(\tfrac{3}{4}, -1\tfrac{1}{8}\right) \]
Choose the style
Why: The symbol is strict.
Test the point (1, 2)
Why: Is two less than two minus three?
\[ 2 < -1 \text{ is false} \]
Shade the other region
Why: The tested point was inside.
Figure (svg): Testing a point to decide which region to shade
\[ \text{dashed curve}; \; \text{shade the region not containing } (1,2) \]
Verify: test a second point in the shaded region
Why: Take (1, −3): is negative three less than negative one? Yes, so that point is a solution and it lies outside the parabola. A confirming test in the region you did shade is worth the ten seconds.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.8 Graphing Quadratic Inequalities §9.8, pp. 548-548
Faded example
A false test sends you to the other side.
Fill in the blanks
Testing the point at one, two gives two is less than -1, which is false, so shade the region not containing it.
Why: A failed test eliminates the tested region, and since there are only two, the other one is selected. That is why one substitution is always enough.
Worked example
The Study Tip's recommendation, when it applies.
\[ \text{Sketch the graph of } y \ge x^2 - 4. \]
Sketch the boundary
Why: Opens up, vertex at nought, negative four.
\[ (0, -4) \]
Choose the style
Why: The symbol includes equality.
Test the origin
Why: Is nought at least negative four?
\[ 0 \ge -4 \text{ is true} \]
Shade its region
Why: The origin is inside the parabola.
Figure (svg): Shading above or below by reading the inequality symbol
\[ \text{solid curve}; \; \text{shade the region containing the origin} \]
Verify: check why the origin was usable here
Why: The origin is not on this curve, since the curve passes through nought, negative four rather than the origin. Whenever the constant term is not nought the origin is off the curve and available as a test point, which makes it the default choice.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.8 Graphing Quadratic Inequalities §9.8, pp. 548-548
Error analysis
The student graphed y less than two x squared minus three x by testing the origin.
Annotate
On: \( \begin{aligned} \text{test } (0,0): \quad 0 &< 2(0)^2 - 3(0) \\ 0 &< 0 \quad \text{false} \\ \text{so shade the region } &\text{not containing the origin} \end{aligned} \)
The Study Tip says the origin is usually good precisely because it is usually off the curve, and the word usually is doing real work. Checking that the test point is not on the boundary takes one substitution and is the same substitution you were going to do anyway.
Elimination
For the inequality y less than two x squared minus three x.
Eliminate the wrong options
Which point can serve as a test point?
Survives elimination: A
Why: A test point must lie strictly off the boundary, and three of these are on it. Checking that first costs the same substitution you were about to perform anyway, so it is effectively free.
Sorting
It is safe when it is not on the curve.
Sort into buckets
Sort each boundary by whether the origin can be used to test it.
The rule is exactly whether the constant term is nought, which can be read off without any substitution at all. Three of the six here fail on that single check.
Socratic
A region contains infinitely many points.
Discussion prompt
Explain why testing a single point decides the shading for a whole region. Then say what would have to be true for that reasoning to fail.
Hint: Ask whether the inequality could change truth value within a region.
Answer:
Within one region, every point is on the same side of the boundary, so the comparison between the point's y and the function's value at that x has the same direction throughout. The truth of the inequality cannot change without the comparison reversing, and that reversal can only happen by crossing the curve — which would mean leaving the region.
The reasoning would fail if the boundary did not actually separate the two cases, for instance if the inequality were not equivalent to a comparison with a single function of x. That is why the method needs the inequality written with y alone on one side, and why the first step of the whole procedure is to get it into that form.
Section
Section 4
Concept
When y is alone on the left, a greater-than symbol means shade above the parabola and a less-than symbol means shade below. No test point is needed.
This describes the same two regions as inside and outside.
Figure (svg): Shading above or below by reading the inequality symbol
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.8 Graphing Quadratic Inequalities §9.8, pp. 549-549 — Method II: Graphing a Quadratic Inequality, and Example 3
Picture it
Read the symbol, shade that side.
Figure (svg): Shading above or below by reading the inequality symbol
For an upward parabola, above is the same as inside; for a downward one it is the same as outside. Using above and below avoids having to keep track of which.
Worked example
This is Example 3 from the textbook.
\[ \text{Sketch the graph of } y \le -x^2 - 5x + 4. \]
Find the vertex
Why: Negative b over two a.
\[ x = -2\tfrac{1}{2} \]
Build the table
Why: Whole numbers either side.
\[ 4, 8, 10, 10\tfrac{1}{4}, 10, 8, 4 \]
Choose the style
Why: The symbol includes equality.
Shade
Why: Less than or equal to means below.
Figure (svg): A downward parabola with the region below it shaded
\[ \text{vertex } \left(-2\tfrac{1}{2}, 10\tfrac{1}{4}\right); \; \text{solid}; \; \text{shade below} \]
Verify: test a point in the shaded region
Why: Take the origin: is nought at most four? Yes, so the origin is a solution, and the origin lies below the curve at x equal to nought where the curve is at four. The reading and a test point agree, which is the sort of confirmation worth doing once per problem.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.8 Graphing Quadratic Inequalities §9.8, pp. 549-549
Translation
Style and side together.
Match the pairs
Why: Every combination of the two independent decisions appears exactly once. Reading them separately — equals part for style, direction for side — reproduces all four without memorising a list.
Worked example
Guided Practice 4 to 6, with no test points.
\[ \text{Describe the graphs of } y < x^2 + 2x - 2, \; y > x^2 - 2x + 3 \text{ and } y \ge 2x^2 + 4x + 2. \]
Take the first
Why: Strict, less than.
Take the second
Why: Strict, greater than.
Take the third
Why: Or-equal, greater than.
Note what was not needed
Why: No substitutions at all.
Figure (svg): Shading above or below by reading the inequality symbol
\[ \text{dashed, below}; \; \text{dashed, above}; \; \text{solid, above} \]
Verify: spot-check the third with a point
Why: The third boundary has vertex at negative one, nought. Testing the origin: is nought at least two? No — and the origin is below that curve at x equal to nought, where the curve is at two. Shading above was right, and the test agrees.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.8 Graphing Quadratic Inequalities §9.8, pp. 549-549
Trap
\[ y - x^2 < 3x \;\Longrightarrow\; \text{less than, so shade below} \]
Read the symbol and shade below immediately
Why: The symbol is less than, and Method II says less than means below.
Method II assumes y stands alone on the left. Here it does not, and rearranging gives y less than x squared plus three x — which does happen to shade below, but only because adding x squared to both sides left the direction unchanged. A step that multiplied by a negative would have reversed it.
\[ y - x^2 < 3x \;\to\; y < x^2 + 3x \;\to\; \text{shade below} \]
Isolate y first, watching for any multiplication by a negative
Why: Only then does the symbol name the region.
Method I needs no such rearrangement, which is why it is worth keeping.
Faded example
When y is already alone.
Fill in the blanks
For y at most the expression, the boundary is solid and the shaded region lies below it.
Why: No arithmetic is involved once y is isolated, which is what makes this method faster. The cost is that it only applies when the inequality is in that form.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Parabola opens | Above the curve is | Below the curve is |
|---|---|---|
| up | inside | outside |
| down | outside | inside |
| either | where y is larger | where y is larger for the curve |
The first two rows disagree, which is why the third row's language is the one the method uses. Above and below mean the same thing regardless of direction, and inside and outside do not.
Socratic
It is stated as a rule.
Discussion prompt
Explain why a greater-than symbol corresponds to the region above the curve. Then say why the rule requires y to be alone on the left.
Hint: What does a larger y mean on a coordinate plane?
Answer:
A larger y-coordinate places a point higher on the plane, so asking for y greater than the function's value at that x is asking for points higher than the curve at that x. Doing that for every x sweeps out exactly the region above the whole curve.
If y is not alone, the left side is some combination whose size does not directly say how high the point is, so the comparison is no longer about vertical position. Worse, isolating y might require multiplying by a negative, which reverses the symbol — so a graph shaded from the unrearranged form could be shaded on precisely the wrong side.
Section
Section 5
Concept
Both methods give the same graph. Reading the symbol is faster when y is already isolated, and testing a point works whatever form the inequality is in.
A test point also catches an error in the boundary sketch.
Figure (svg): Two columns comparing the test-point method with the above-or-below method
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.8 Graphing Quadratic Inequalities §9.8, pp. 548-549 — Methods I and II presented as alternatives
Picture it
Speed against generality.
Figure (svg): Two columns comparing the test-point method with the above-or-below method
They are not rival methods so much as the same decision reached two ways. Using one and checking with the other is the most reliable habit.
Worked example
One graph, two routes.
\[ \text{Graph } y > x^2 - 4 \text{ by reading, then confirm by testing.} \]
Read the symbol
Why: Strict, greater than.
Choose a test point
Why: The origin is off the curve.
\[ (0, 0) \]
Substitute
Why: Is nought greater than negative four?
Compare
Why: The origin is above the curve.
Figure (svg): Two columns comparing the test-point method with the above-or-below method
\[ \text{dashed}; \; \text{shade above, containing the origin} \]
Verify: say what a disagreement would have meant
Why: If the reading had said above and the test had landed in the region below, one of the two would be wrong — most likely the sketch of the boundary, since that is the step with the most arithmetic in it. Agreement between two independent routes is worth more than care on either one alone.
Hypothesis
Given two x squared minus y at least six.
Predict first
What is the safest first move?
Correct: Sketch the boundary and test a point in the original inequality.
\[ 2x^2 - y \ge 6 \;\to\; -y \ge 6 - 2x^2 \;\to\; y \le 2x^2 - 6 \]
Why: Method II needs y alone on the left, and here it is not, so reading the symbol directly would be guessing. Isolating y requires dividing by a negative, which reverses the symbol to y at most two x squared minus six — so the correct shading is below, and both of the reading options were premature. Testing a point in the inequality as given avoids the reversal entirely and is why Method I remains worth knowing.
Worked example
An inequality that is not solved for y.
\[ \text{Graph } x^2 - y < 4 \text{ without isolating } y \text{ first.} \]
Sketch the boundary
Why: The curve where the two sides are equal.
\[ y = x ^{2} - 4 \]
Choose the style
Why: The symbol is strict.
Test the origin
Why: Nought minus nought is nought.
\[ 0 < 4 \text{ is true} \]
Shade its region
Why: The origin is above the curve.
Figure (svg): Two columns comparing the test-point method with the above-or-below method
\[ \text{dashed}; \; \text{shade the region containing the origin} \]
Verify: isolate y and compare
Why: Rearranging gives negative y less than four minus x squared, and dividing by negative one reverses the symbol to y greater than x squared minus four — which shades above, as the test found. The reversal is exactly the trap Method I sidesteps.
Trap
\[ x^2 - y < 4 \;\to\; -y < 4 - x^2 \;\to\; y < x^2 - 4 \]
Multiply through by negative one and keep the symbol
Why: Only the signs changed, so the symbol looked safe.
Multiplying an inequality by a negative reverses it. Testing the origin settles it: nought minus nought is less than four, so the origin is a solution — but it does not satisfy y less than x squared minus four, since nought is not less than negative four.
\[ -y < 4 - x^2 \;\to\; y > x^2 - 4 \]
Reverse the symbol whenever both sides are multiplied or divided by a negative
Why: The rule from Lesson 6.2 has not changed.
Or avoid the issue altogether by testing a point in the original inequality.
Sorting
Is y alone on the left?
Sort into buckets
Sort each inequality by whether the symbol can be read directly.
The last one needs only a subtraction to isolate y, with no sign reversal, so it is quick either way. Two of the others need a division by a negative, which is where the reversals hide.
Faded example
Isolating y flips the symbol.
Fill in the blanks
Dividing by negative one reverses the symbol from less than to greater than, so the shaded region lies above the boundary.
Why: Dividing both sides by negative one reverses the direction, turning a less-than into a greater-than. Keeping the symbol unchanged would shade the wrong region entirely, which a single test point would have exposed.
Socratic
One of them is always faster.
Discussion prompt
Say what would be lost by learning only the reading method. Then say what would be lost by learning only the test-point method.
Hint: Think about generality and about speed.
Answer:
Learning only the reading method means being stuck whenever y is not isolated, and worse, being tempted to read the symbol anyway — which gives a confidently wrong answer whenever isolating y would have required a sign reversal. It also gives no independent check on the boundary sketch.
Learning only the test-point method costs a substitution on every problem, which is minor, and gives no intuition for why the regions correspond to the symbols, which is not. The reading method makes the connection between algebra and picture visible, and that connection is what the whole chapter has been building towards — so the two are worth keeping for different reasons.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Decision | Decided by | Example |
|---|---|---|
| Dashed or solid | whether the symbol includes equality | at most gives a solid curve |
| Which region | the direction of the symbol, or a test point | greater than shades above |
| Where the boundary sits | the corresponding equation | sketch it from its vertex and a table |
Three independent decisions, and each can be got right or wrong on its own. Working through them in order is what keeps a wrong boundary from being confused with wrong shading.
Pattern
To sketch the graph of any quadratic inequality in two variables, these five moves cover it.
Step five costs one substitution and catches errors in every one of the four steps before it, including a mis-sketched boundary.
OpenStax Intermediate Algebra 2e, §9.8 Solve Quadratic Inequalities §9.8
Check
Compare the two heights.
Check your understanding
Is the point (4, -3) inside or outside the parabola y = x squared - 3x - 3?
Answer: A
Why: At x equal to four the curve is at sixteen minus twelve minus three, which is one. The point's height of negative three is below that, and below an upward parabola is outside it.
Check
Two decisions from one symbol.
Check your understanding
How should the graph of y at most x squared minus 1 be drawn?
Answer: A
Why: The symbol includes equality, so the boundary is solid, and it is a less-than direction, so the region below is shaded.
Check
Check the test point is usable.
Check your understanding
Why can the origin not be used to test y less than 2x squared - 3x?
Answer: A
Why: The function has no constant term, so it passes through the origin, and a point on the boundary lies in neither region and settles nothing.
Real world
This is the flashlight question from the lesson opener. A flashlight has a parabolic reflector, and the bulb sits inside the curve so that light bounces off it into a beam.
Discussion prompt
Suppose a reflector's cross-section follows y equals x squared, with distances in centimetres. Describe the region occupied by the reflector's interior as an inequality, and say where a bulb placed at (0, 0.25) sits relative to it.
Hint: Interior means above, for an upward curve.
Answer:
\[ y > x^2 \quad \text{describes the interior} \]
The bulb at nought, a quarter has y equal to 0.25 and the curve at x equal to nought is at nought, so the bulb is above the curve and therefore inside the reflector — which is where it must be if the light is to bounce off the surface.
Whether the boundary is included is a real question here rather than a formality: with a strict symbol the reflective surface itself is excluded from the interior, which is right if you are describing where the air and the bulb are, and wrong if you are describing the piece of metal. Which symbol to use depends on what is being described, and that is a modelling decision the mathematics cannot make.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
For x squared minus y less than 4, which region should be shaded?
Correct: Above the boundary.
\[ x^2 - y < 4 \;\to\; y > x^2 - 4 \]
Why: The symbol cannot be read directly because y is not alone on the left. Isolating it gives negative y less than four minus x squared, and dividing by negative one reverses the direction to y greater than x squared minus four — so the region above is shaded. Testing the origin confirms it without any rearranging at all: nought minus nought is less than four, so the origin is a solution, and the origin lies above the curve which passes through nought, negative four. The first option is the trap, and it comes from applying Method II to an inequality that is not in the form Method II requires.
Explain it
They tested the origin on y less than two x squared minus three x and concluded nothing useful.
Discussion prompt
In no more than four sentences, explain why that test point failed them and how to spot the problem in advance. Then suggest a point that does work.
Hint: Where does that curve pass?
Answer:
A usable answer: the function has no constant term, so its graph passes through the origin, and a point on the boundary is in neither region. A test point has to be strictly off the curve, or it cannot choose between the two sides.
You can spot it in advance by checking whether the constant term is nought — if it is, the curve goes through the origin. Something like the point (1, 2) works instead: two is not less than negative one, so the region containing it is not shaded.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The boundary is fixed by Lesson 9.4's method: vertex first, then a table centred on it. Dashed or solid is fixed by asking only whether the symbol includes equality. The region is fixed by reading the direction when y is alone and testing a point when it is not. The last is fixed by remembering that dividing by a negative reverses the symbol, or by testing in the original form and sidestepping the issue. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page write the four kinds of quadratic inequality and beside each write its boundary style and its region in two words. Underneath, draw one parabola and mark three points on the page — one clearly inside, one clearly outside and one close to the curve — then classify all three by substitution rather than by eye, showing the two heights you compared each time. In the middle, graph y less than two x squared minus three x completely: find the vertex, build a table, draw the boundary dashed, mark the point you tested, write the substitution and its truth value, and shade the region your test selected. Beneath that, graph y at most negative x squared minus five x plus four using the reading method instead, with a solid curve and the region below shaded, then confirm it with one test point. In the lower corner, write an inequality where y is not isolated, isolate it showing the sign reversal, and note that a test point in the original would have avoided the reversal entirely. Finally, in the margin, write the check for whether the origin is usable as a test point.
Your confirming test point should always land in the region you shaded. If it does not, work backwards through the three decisions — boundary, style, region — since only one of them is usually wrong.
Recap
Five things, and the last two are two routes to the same decision.
| If the question says | Your first move is |
|---|---|
| Is this point inside the parabola? | Substitute its x and compare heights |
| The symbol is < or > | Draw the boundary dashed |
| The symbol is at most or at least | Draw the boundary solid |
| y is alone on the left | Read the direction and shade that side |
| y is not alone | Test a point in the inequality as given |
That completes Chapter 9. Chapter 10 turns to the expressions themselves rather than their graphs: adding, multiplying and above all factoring polynomials, which will give a second exact way of solving the quadratic equations this chapter has been graphing.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.8 Graphing Quadratic Inequalities §9.8, pp. 547-552 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Algebra 1 — $55/session, free consultation.