Using a graph to find or check the solutions of a quadratic equation. Includes the connection between x-intercepts and roots, rewriting an equation into standard form before graphing, estimating solutions from a sketch and confirming them algebraically, reading the number of solutions off a graph, and applying the method to a bridge cable model.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 9 — Quadratic Equations and Functions
Solving Quadratic Equations by Graphing
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 9.5 Solving Quadratic Equations by Graphing §9.5, pp. 526-532 — the lesson these objectives are drawn from
Warm-up
Lesson 9.4 drew parabolas. This lesson asks a question about them that turns out to be an equation.
Discussion prompt
The graph of y equals x squared minus four crosses the x-axis at two points. Where are they, and what equation do those two numbers solve?
Hint: At a crossing point the height is nought.
Answer:
\[ x^2 - 4 = 0 \;\Longrightarrow\; x = \pm 2 \]
The crossings are at two and negative two, and those are exactly the solutions of x squared minus four equals nought. Asking where a graph meets the axis and asking which values make the function nought are the same question asked in two languages.
Concept
The x-intercepts of the graph of y equals a x squared plus b x plus c are the solutions of the related equation a x squared plus b x plus c equals nought. At such a point y is nought.
roots of a quadratic equation — The solutions of a quadratic equation in one variable. They are the x-intercepts of the graph of the related quadratic function.
Solutions and roots are two names for the same numbers.
Figure (svg): A parabola crossing the x-axis at the solutions of the related equation
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.5 Solving Quadratic Equations by Graphing §9.5, pp. 526-527
Section
Section 1
Concept
An x-intercept is the x-coordinate of a point where a graph crosses the x-axis, and at that point y is nought. So the x-intercepts of a quadratic function are precisely the values that make it nought.
\[ y = ax^2 + bx + c = 0 \quad \text{at each x-intercept} \]
The solutions are also called the roots.
Figure (svg): A parabola crossing the x-axis at the solutions of the related equation
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.5 Solving Quadratic Equations by Graphing §9.5, pp. 526-526 — the statement connecting x-intercepts with the related equation, and Example 1
Picture it
Height nought at each.
Figure (svg): A parabola crossing the x-axis at the solutions of the related equation
Nothing has to be solved to see how many solutions there are. Counting crossings answers that before any of the numbers is known.
Worked example
This is Example 1 from the textbook.
\[ \text{The graph of } y = \tfrac{1}{2}x^2 - 8 \text{ is given. Estimate the solutions of } \tfrac{1}{2}x^2 - 8 = 0. \]
Find where the curve meets the axis
Why: Read the crossing points.
\[ (-4, 0) \text{ and } (4, 0) \]
Take the x-coordinates
Why: Those are the candidate solutions.
\[ -4 \text{ and } 4 \]
Check the first
Why: Substitute negative four.
\[ \tfrac{1}{2}(16) - 8 = 0 \]
Check the second
Why: Substitute four.
\[ \tfrac{1}{2}(16) - 8 = 0 \]
Figure (svg): A parabola with x-intercepts at four and negative four
\[ x = 4 \text{ and } x = -4 \]
Verify: note why the two are symmetric
Why: This function has no x term, so its axis of symmetry is the y-axis and the two intercepts must be a number and its negative. Whenever b is nought the roots come as a matched pair, which is a useful sanity check on a reading.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.5 Solving Quadratic Equations by Graphing §9.5, pp. 526-526
Matching
Which intercept answers which question.
Match the pairs
Why: Only the first row is about solving. The last is worth noticing: since the roots are symmetric about the axis, the axis always sits at their midpoint, which gives a way of finding one root from the other.
Worked example
Guided Practice 1, where one crossing sits at nought.
\[ \text{From the graph of } y = 2x^2 - 4x, \text{ estimate the solutions of } 2x^2 - 4x = 0. \]
Read the crossings
Why: The curve meets the axis twice.
\[ x = 0 \text{ and } x = 2 \]
Check the first
Why: Substitute nought.
\[ 0 - 0 = 0 \]
Check the second
Why: Substitute two.
\[ 8 - 8 = 0 \]
Note the missing constant
Why: There is no c term.
Figure (svg): The solution to Worked example a root at the origin shown as a ladder of expressions, one row per algebraic move
\[ x = 0 \text{ and } x = 2 \]
Verify: connect the missing constant to the root at nought
Why: The y-intercept is c, which here is nought, so the curve passes through the origin — and the origin is on the x-axis, making nought automatically a root. A quadratic with no constant term always has nought as one of its solutions.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.5 Solving Quadratic Equations by Graphing §9.5, pp. 526-526
Trap
The graph of y equals x squared minus x minus two crosses the y-axis at negative two.
Report negative two as a solution
Why: It is an intercept, and intercepts are solutions.
Only x-intercepts are solutions. The y-intercept is where x is nought, not where y is nought, and substituting negative two gives four plus two minus two, which is four rather than nought.
\[ \text{x-intercepts } -1 \text{ and } 2 \;\Longrightarrow\; \text{solutions } -1 \text{ and } 2 \]
Look for crossings of the horizontal axis only
Why: Solving means setting y to nought, which is a horizontal line.
The y-intercept is useful for sketching and never for solving.
Faded example
State what is true at a crossing.
Fill in the blanks
At an x-intercept the value of y is zero, so the x-coordinate makes the expression equal zero.
Why: The equation asks which x values make the expression nought, and the graph shows exactly where its height is nought. The two questions are identical, which is why the method works at all.
Elimination
From the graph of a quadratic function.
Eliminate the wrong options
Which coordinate should be reported as a solution?
Survives elimination: A
Why: Solving asks for values of x, so an x-coordinate is what gets reported. Option B is a real trap on a test, because the point really is on the graph and its y-coordinate really is nought — it just says nothing.
Socratic
The graph and the equation look unrelated.
Discussion prompt
Explain the connection between solving an equation and finding where a curve meets an axis. Then say what solving a different equation, such as the function equal to three, would look like on the graph.
Hint: Ask what the equals sign is asking for.
Answer:
The function assigns a height to every x, and the equation asks which x values give a height of nought. The x-axis is the set of all points at height nought, so the x values that satisfy the equation are precisely the ones where the curve touches that line.
Setting the function equal to three would ask where the curve reaches a height of three, which is where it crosses the horizontal line y equals three. That line can miss the curve, touch it once or cut it twice, exactly as the axis can — so the whole method generalises, and the special role of nought is only that standard form puts it there.
Section
Section 2
Concept
Before graphing, write the equation in standard form with nought on one side. Then graph the related function, whose x-intercepts are the solutions.
Graphing the equation as first written answers a different question.
Figure (svg): Rewriting an equation into standard form before graphing
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.5 Solving Quadratic Equations by Graphing §9.5, pp. 527-527 — the Estimating Solutions by Graphing steps and Example 2
Picture it
The zero has to be on one side.
Figure (svg): Rewriting an equation into standard form before graphing
The function to plot is whatever ends up opposite the nought. That is the whole reason the rearrangement comes first rather than last.
Worked example
This is Example 2 from the textbook.
\[ \text{Use a graph to estimate the solutions of } x^2 - x = 2. \]
Write in standard form
Why: Subtract two from each side.
\[ x ^{2} - x - 2 = 0 \]
Name the related function
Why: The expression opposite the nought.
\[ y = x ^{2} - x - 2 \]
Sketch it
Why: A parabola opening up.
\[ vertex at x = \tfrac{1}{2} \]
Read the x-intercepts
Why: Where it crosses.
\[ -1 \text{ and } 2 \]
Figure (svg): A parabola crossing the x-axis at the solutions of the related equation
\[ x = -1 \text{ and } x = 2 \]
Verify: check in the original equation
Why: At negative one the left side is one plus one, which is two, and at two it is four minus two, also two. Both match the right side of the equation as first written, which confirms the rearrangement as well as the reading.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.5 Solving Quadratic Equations by Graphing §9.5, pp. 527-527
Faded example
Zero on one side.
Fill in the blanks
x^2 - x = 6 \;\to\; x^2 - x - 6 = 0 \;\to\; \text6 y = x^2 - x - ___
Why: The function to graph is whatever sits opposite the nought after rearranging. Graphing the left side alone would find where it equals nought, which is a different question with different answers.
Worked example
Guided Practice 2 and 3.
\[ \text{Use a graph to estimate the solutions of } x^2 - x = 6. \]
Write in standard form
Why: Subtract six from each side.
\[ x ^{2} - x - 6 = 0 \]
Graph the related function
Why: Opens up, vertex at a half.
\[ y = x ^{2} - x - 6 \]
Read the intercepts
Why: Two crossings.
\[ -2 \text{ and } 3 \]
Check both
Why: Substitute into the original.
\[ 6 \text{ each time} \]
Figure (svg): The solution to Worked example a second rearrangement shown as a ladder of expressions, one row per algebraic move
\[ x = -2 \text{ and } x = 3 \]
Verify: compare with the previous example
Why: The same left side with six instead of two moved the roots from negative one and two out to negative two and three. Raising the right side lowers the graph, so the crossings spread apart, and the midpoint stays at a half in both cases.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.5 Solving Quadratic Equations by Graphing §9.5, pp. 527-527
Error analysis
The student was solving x squared minus x equals two by graphing.
Annotate
On: \( \begin{aligned} \text{graph } y &= x^2 - x \\ \text{x-intercepts: } x &= 0 \text{ and } x = 1 \\ \text{so the solutions are } &0 \text{ and } 1 \end{aligned} \)
This is the single commonest error in the lesson, and it produces perfectly reasonable-looking numbers. The check is what catches it, which is why substituting into the original equation is worth the twenty seconds it costs.
Sorting
Standard form means zero on one side.
Sort into buckets
Sort each equation by whether it can be graphed as it stands.
Half of them need a rearrangement, and in every one of those cases graphing the left side alone would give wrong answers that look entirely plausible.
Elimination
To solve x squared plus three equals four x.
Eliminate the wrong options
Which related function has the right x-intercepts?
Survives elimination: A
Why: Subtracting four x from both sides gives the correct related function, whose roots are one and three. Option D is the sign slip worth watching, since it produces two perfectly tidy roots that are both wrong.
Socratic
Two graphs instead of one.
Discussion prompt
Suppose you graphed y equals x squared minus x and y equals two on the same axes. Where would the solutions be, and why does the textbook rearrange instead?
Hint: Ask where the two graphs meet.
Answer:
The solutions would be the x-coordinates of the points where the parabola meets the horizontal line, since those are the x values at which the two sides are equal. That method works perfectly well and is what a graphing calculator's intersect feature does.
Rearranging is preferred because it reduces every problem to the same question — where does a curve meet the x-axis — rather than a different horizontal line each time. It also connects directly to the standard form used by the quadratic formula in Lesson 9.6, so the same rearrangement serves both methods and does not have to be learnt twice.
Section
Section 3
Concept
A reading off a graph is an estimate. Substituting it into the original equation turns it into a confirmed solution, or reveals that it was only close.
Whole-number roots are usually exact; others rarely are.
Figure (svg): Checking two estimated roots by substitution
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.5 Solving Quadratic Equations by Graphing §9.5, pp. 527-527 — the algebraic check following Example 2
Picture it
Both sides have to match.
Figure (svg): Checking two estimated roots by substitution
Checking in the original equation rather than the rearranged one tests the rearrangement too, which is where the previous section's error hid.
Worked example
The check from Example 2, written out.
\[ \text{Check that } -1 \text{ and } 2 \text{ solve } x^2 - x = 2. \]
Substitute negative one
Why: Mind the brackets.
\[ (-1) ^{2} - (-1) \]
Simplify
Why: One plus one.
\[ 2 \;\checkmark \]
Substitute two
Why: Square first.
\[ 2 ^{2} - 2 \]
Simplify
Why: Four minus two.
\[ 2 \;\checkmark \]
Figure (svg): Checking two estimated roots by substitution
\[ (-1)^2 - (-1) = 2, \qquad 2^2 - 2 = 2 \]
Verify: notice which equation was used
Why: The check used the original equation, not the rearranged one, so it confirms the subtraction of two as well as the graph reading. Checking in the rearranged version would have missed an error made during the rearrangement.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.5 Solving Quadratic Equations by Graphing §9.5, pp. 527-527
Faded example
Negative values need care.
Fill in the blanks
x = -1: \quad (-1)^2 - (-1) = 1 + 1 = 2
Why: Both the squaring and the subtraction involve a negative, and the brackets keep them straight. Writing it without brackets would give negative one minus one, which is negative two and would wrongly reject a correct solution.
Worked example
A root that is not a whole number.
\[ \text{Estimate the positive solution of } x^2 - 3 = 0 \text{ from a graph, then check.} \]
Read the intercept
Why: Between one and two, nearer two.
\[ \text{about } 1.7 \]
Substitute
Why: 1.7 squared is 2.89.
\[ 2.89 - 3 = -0.11 \]
Interpret
Why: Close to nought but not nought.
Solve exactly
Why: Take square roots.
\[ x = \sqrt{3} \approx 1.732 \]
Figure (svg): A graph giving an estimate rather than an exact value
\[ x = \sqrt{3} \approx 1.73 \]
Verify: say what the check actually told you
Why: The substitution gave a small non-zero value, which shows the estimate was close but not exact rather than that it was wrong. That is the normal outcome for an irrational root, and it is why graphing is described as estimating solutions rather than finding them.
Trap
The curve seems to cross a little past one and a half, so the solution is 1.5.
Report the graph reading as the answer
Why: The graph is the method being used, so its reading must be the answer.
Substituting 1.5 gives 2.25 minus three, which is negative 0.75 — a long way from nought. A sketch cannot resolve much better than the nearest half unit, and here the true root is about 1.73.
\[ x = \sqrt{3} \approx 1.73 \]
Use the graph for the count and the rough location, then confirm or refine algebraically
Why: The two methods answer different parts of the question.
Whole-number readings usually check out exactly; anything between gridlines rarely does.
Hypothesis
Substituting 1.7 into x squared minus three gives -0.11.
Predict first
What should be concluded?
Correct: The estimate is close but the true root is not exactly 1.7.
\[ 1.7^2 - 3 = -0.11 \qquad 1.73^2 - 3 = -0.0071 \]
Why: A result near nought says the value is near a root, and a result exactly nought says it is one. Since the true root is the root of three, about 1.732, no decimal with one place could check exactly. The size of the discrepancy is a rough measure of how far off the estimate is, and here a small negative value also says the estimate is on the low side of the root.
Sorting
For solving x squared minus x equals two.
Sort into buckets
Sort each candidate by whether it is a sound thing to check against.
A check is only worth doing against something that could not have inherited the error. That rules out every line you wrote yourself after the first.
Socratic
It rarely gives exact answers.
Discussion prompt
Name two things a graph tells you about a quadratic equation that the algebra of Lesson 9.2 does not. Then say what it is poor at.
Hint: Think about what you see before computing anything.
Answer:
First, it shows how many solutions there are at a glance, by counting crossings — no work is needed to know whether there will be two, one or none. Second, it shows roughly where they lie and how the function behaves between and beyond them, which matters when a situation asks not just for the roots but for where the quantity is positive or largest.
It is poor at precision. A hand sketch resolves to about half a gridline, so an irrational root can only be located approximately, and two roots very close together may look like one or like none. That is why the textbook calls this estimating solutions, and why Lesson 9.6 supplies a formula that is exact for every case.
Section
Section 4
Concept
A parabola can meet the x-axis twice, once, or not at all, and the count is the number of real solutions of the related equation. This is the same three-case split as in Lesson 9.2.
The vertex's position relative to the axis decides which.
Figure (svg): Parabolas crossing the axis twice, once and not at all
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.5 Solving Quadratic Equations by Graphing §9.5, pp. 527-527 — the statement that the roots are the x-intercepts, if any
Picture it
The three possible pictures.
Figure (svg): Parabolas crossing the axis twice, once and not at all
For an upward parabola the vertex below the axis gives two roots, on it gives one, and above it gives none. A downward parabola works the same way upside down.
Worked example
Three functions, three answers.
\[ \text{How many real roots have } x^2 - 1 = 0, \; x^2 = 0 \text{ and } x^2 + 1 = 0? \]
Take the first
Why: Vertex at nought, negative one, below the axis.
Take the second
Why: Vertex at the origin, on the axis.
Take the third
Why: Vertex at nought, one, above the axis.
State the counts
Why: In the same order.
Figure (svg): Parabolas crossing the axis twice, once and not at all
\[ \pm 1; \quad 0; \quad \text{no real solution} \]
Verify: compare with Lesson 9.2's rule
Why: Rearranged, these are x squared equal to one, nought and negative one, which the earlier rule sorts as two, one and no solutions. The graph and the algebra give the same counts, as they must, since they are answering the same question.
Sorting
Vertex position and direction decide.
Sort into buckets
Sort each equation by its number of real solutions.
Two of the six have no real roots, which is a higher proportion than most exercise sets suggest. Assuming a pair always exists is a habit worth breaking early.
Worked example
Reading the count from the vertex alone.
\[ \text{Without graphing, how many roots has } x^2 - 2x + 5 = 0? \]
Find the axis
Why: Negative b over two a.
\[ x = 1 \]
Find the vertex height
Why: Substitute one.
\[ 1 - 2 + 5 = 4 \]
Note the direction
Why: The leading coefficient is positive.
Conclude
Why: Lowest point four above the axis.
Figure (svg): Parabolas crossing the axis twice, once and not at all
\[ \text{vertex } (1, 4), \text{ opens up} \;\Longrightarrow\; \text{no real roots} \]
Verify: test a value to confirm
Why: At x equal to nought the function gives five and at x equal to two it gives five again, both positive, and the minimum is four. The function never reaches nought, so the equation has no real solution and no amount of searching would find one.
Trap
\[ x^2 - 2x + 5 = 0 \]
Look for two solutions, since quadratics have two
Why: Every example so far produced a pair.
The graph never reaches the axis: its lowest point is four units above it. Searching for two roots here is looking for something that does not exist, and a sketch settles it in seconds.
The vertex is at (1, 4) and the parabola opens up, so there are no real roots.
Locate the vertex relative to the axis before hunting for roots
Why: That decides the count immediately.
Lesson 9.7 turns this observation into a single number that predicts the count without any graphing at all.
Prediction
A parabola touching the axis once.
Predict first
For an equation with exactly one real solution, where is the vertex?
Correct: Exactly on the x-axis.
\[ y = x^2 \text{ has vertex } (0,0) \text{ and exactly one root} \]
Why: The curve turns at its vertex, so if the vertex is off the axis the curve either crosses twice on its way past or never reaches the axis at all. Only when the turning point is exactly on the axis does the curve touch it once and turn back, which is the graphical picture of the two roots of Lesson 9.2 coinciding at nought.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Opens up, vertex is | Roots | Because |
|---|---|---|
| below the axis | two | the arms rise back through the axis on both sides |
| on the axis | one | the curve touches and turns |
| above the axis | none | the whole curve stays above the axis |
A downward parabola gives the same three cases with above and below swapped. Combining the direction with the vertex height is enough to count the roots of any quadratic without solving it.
Socratic
A wavy curve could.
Discussion prompt
Explain why a parabola meets a horizontal line at most twice. Then say what that implies about the number of solutions a quadratic equation can have.
Hint: The curve turns exactly once.
Answer:
A parabola falls, turns once at its vertex, and then rises — or the reverse. Each of those two stretches is heading steadily in one direction, so it can pass any given height at most once, which caps the total number of crossings at two.
So a quadratic equation has at most two real solutions, and this is not a fact about the formula but about the shape. A curve that turned twice could meet a line three times, which is why cubic equations can have three solutions — and Lesson 10.8 meets exactly that situation.
Section
Section 5
Concept
When a quadratic function models a situation, solving for a particular height means finding where the graph meets a horizontal line. Rearranging turns that into an x-intercept problem.
The interpretation step is where most marks are lost.
Figure (svg): The suspension cable of a bridge modelled as a parabola
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.5 Solving Quadratic Equations by Graphing §9.5, pp. 528-528 — Example 3, Points on a Parabola, on the Golden Gate Bridge
Picture it
Two towers, one curve.
Figure (svg): The suspension cable of a bridge modelled as a parabola
The horizontal distance is measured from the middle, so each intercept locates one tower and the answer is the gap between them rather than either value on its own.
Worked example
This is Example 3 from the textbook.
\[ \text{The cable follows } y = 0.000112x^2 + 8 \text{ and meets the towers at } 500 \text{ feet. Find the span.} \]
Set the height
Why: The towers meet the cable at five hundred feet.
\[ 0.000112 x ^{2} + 8 = 500 \]
Write in standard form
Why: Subtract five hundred.
\[ 0.000112 x ^{2} - 492 = 0 \]
Find the intercepts
Why: From a graphing calculator.
\[ about \pm 2100 \]
Interpret
Why: Each tower is that far from the middle.
\[ 2100 + 2100 = 4200 \]
Figure (svg): The suspension cable of a bridge modelled as a parabola
\[ x \approx \pm 2100 \;\Longrightarrow\; \text{span} \approx 4200 \text{ feet} \]
Verify: check one intercept in the model
Why: Substituting 2100 gives 0.000112 times four million four hundred and ten thousand, which is about four hundred and ninety-four, plus eight — about five hundred and two feet. That is within a couple of feet of the stated height, which is as close as a rounded intercept can get.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.5 Solving Quadratic Equations by Graphing §9.5, pp. 528-528
Faded example
The towers are at five hundred feet.
Fill in the blanks
0.000112x^2 + 8 = 500 \;\to\; 0.000112x^2 - 492 = 0 \;\to\; x \approx \pm 2100
Why: The eight is the height of the cable's lowest point above the road, so subtracting it from five hundred leaves the rise from that low point to the towers. Forgetting it would change the answer by several feet.
Worked example
Guided Practice 4, the Royal Gorge Bridge.
\[ \text{With } y = 0.0007748x^2 \text{ and towers at } 150 \text{ feet, find the span.} \]
Set the height
Why: A hundred and fifty feet.
\[ 0.0007748 x ^{2} = 150 \]
Isolate the squared term
Why: Divide by the coefficient.
\[ x^2 \approx 193\,598 \]
Take square roots
Why: Both signs.
\[ x \approx \pm 440 \]
Double the distance
Why: Two towers, one each side.
\[ \text{about } 880\text{ feet} \]
Figure (svg): The suspension cable of a bridge modelled as a parabola
\[ x \approx \pm 440 \;\Longrightarrow\; \text{span} \approx 880 \text{ feet} \]
Verify: check the substitution
Why: 0.0007748 times 440 squared is 0.0007748 times a hundred and ninety-three thousand six hundred, which is about a hundred and fifty. The model returns the tower height at the intercept, confirming both the arithmetic and the reading.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.5 Solving Quadratic Equations by Graphing §9.5, pp. 528-528
Trap
\[ x \approx 2100 \]
Report twenty-one hundred feet as the distance between the towers
Why: That is the number the graph gave.
Twenty-one hundred feet is the distance from the middle of the bridge to one tower. The towers are on opposite sides of the middle, so the distance between them is twice that.
\[ 2100 + 2100 = 4200 \text{ feet} \]
Read what x measures before interpreting the intercepts
Why: Here it is distance from the middle, not from a tower.
Writing down what the variable means before solving is what prevents this entirely.
Elimination
For the bridge model, with x measured from the middle.
Eliminate the wrong options
What does the intercept at about 2100 represent?
Survives elimination: A
Why: The variable was defined as horizontal distance from the middle, so each intercept locates one tower relative to that midpoint. Reading the definition of the variable before interpreting the answer is what separates options A and B.
Prediction
The same cable equation, towers at 600 feet instead of 500.
Predict first
What happens to the distance between the towers?
Correct: It grows, because the cable reaches that height further out.
\[ 500 \text{ ft}: \; \text{span } 4200 \qquad 600 \text{ ft}: \; \text{span about } 4600 \]
Why: The cable rises as it moves away from the middle, so a greater height is reached at a greater horizontal distance. Setting the model to six hundred gives x squared about five million two hundred and eighty-six thousand, so x is about two thousand three hundred, and the span grows to roughly four thousand six hundred feet. The relationship is not proportional, though: a twenty per cent taller tower buys only about a ten per cent longer span, because the height depends on the square of the distance.
Socratic
Earlier examples were sketched by hand.
Discussion prompt
Say why this model is awkward to sketch by hand. Then say what a calculator does and does not add to the method.
Hint: Look at the scale of the numbers.
Answer:
The x values run into the thousands while the coefficient is about a ten-thousandth, so a hand sketch would need a scale on which the interesting part of the curve is either invisible or off the page. Choosing a sensible window is most of the difficulty, and a calculator lets you try several in seconds.
What the calculator adds is precision and speed of plotting; what it does not add is any change to the method, which is still rearrange, graph, read the intercepts and interpret. It also does not decide whether the answer is one intercept or the gap between two, which is the step that actually answers the question and the one no tool will do for you.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Graphing | Square roots (9.2) | |
|---|---|---|
| Works when | always | there is no x term |
| Gives answers that are | estimates | exact |
| Shows the number of roots | yes, by counting crossings | yes, from the sign of d |
Graphing is more general and less precise; square roots are exact but only for one shape of equation. Lesson 9.6 supplies a method that is both general and exact.
Pattern
To estimate the solutions of a quadratic equation by graphing, these five moves cover it.
Step one is the one that is skipped, and skipping it produces plausible wrong answers rather than obvious ones — which is exactly why step five exists.
OpenStax Elementary Algebra 2e, §10.5 Graphing Quadratic Equations in Two Variables §10.5
Check
Rearrange first.
Check your understanding
To solve x squared minus x equals 6 by graphing, which function should you graph?
Answer: A
Why: Subtracting six from both sides puts the equation in standard form, and the expression opposite the nought is the function to graph. Its intercepts are negative two and three.
Check
Count the crossings.
Check your understanding
A parabola opens up and its vertex is at (2, 3). How many real roots has the related equation?
Answer: A
Why: The lowest point is three units above the x-axis and the curve rises from there, so it never reaches the axis and there are no real solutions.
Check
Read what x measures.
Check your understanding
A cable model gives x-intercepts of about -2100 and 2100, with x the distance from the middle. How far apart are the towers?
Answer: A
Why: Each tower is about twenty-one hundred feet from the middle, and they stand on opposite sides, so the span is the sum of the two distances.
Real world
This is the Golden Gate Bridge question from the lesson opener. The main cables follow y equals 0.000112 x squared plus eight, with x measured horizontally from the middle and y measured up from the road.
Discussion prompt
The cables meet the towers 500 feet above the road. Find how far apart the towers are, and say what the eight in the model represents. Then do the same for the Royal Gorge Bridge, whose cables follow y equals 0.0007748 x squared with towers at 150 feet.
Hint: Each intercept locates one tower.
Answer:
\[ 0.000112x^2 + 8 = 500 \;\Longrightarrow\; x \approx \pm 2100 \;\Longrightarrow\; \text{span} \approx 4200 \text{ ft} \]
The eight is the y-intercept, which here is the height of the lowest point of the cables above the roadway at the middle of the bridge — the cables do not touch the road.
The Royal Gorge model has no constant, so its cables reach the road at the midpoint, and setting it to a hundred and fifty gives intercepts of about plus and minus four hundred and forty feet, a span of about eight hundred and eighty. Both answers required doubling an intercept, and both required reading what x was defined to measure before interpreting anything.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
To solve x squared plus 3 equals 4x by graphing, what should you graph?
Correct: y = x squared - 4x + 3, and read its x-intercepts.
\[ x^2 + 3 = 4x \;\to\; x^2 - 4x + 3 = 0 \;\to\; x = 1, \; 3 \]
Why: Standard form comes first, so four x is subtracted from both sides, leaving the expression opposite the nought as the function to graph; its intercepts are one and three, and both check in the original equation since one plus three is four and nine plus three is twelve. Graphing the left side alone would ask where x squared plus three equals nought, which never happens. Adding four x instead of subtracting gives roots of negative one and negative three, two tidy numbers that are both wrong — and that is exactly why the check is done against the equation as first written rather than the rearranged one.
Explain it
They graphed the left side of x squared minus x equals two and got nought and one.
Discussion prompt
In no more than four sentences, explain what went wrong and give them the rule for what to graph. Then tell them the check that would have caught it.
Hint: Where is the nought?
Answer:
A usable answer: graphing the left side alone finds where it equals nought, but the question asked where it equals two. Subtract the two first, then graph what is left opposite the nought — that is y equals x squared minus x minus two, whose intercepts are negative one and two.
The check is to substitute back into the equation as it was first written. Nought squared minus nought is nought, not two, which shows the answer immediately and points at the missing rearrangement.
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: Rearranging is fixed by making the nought the first thing you write. Reading is fixed by treating a between-gridlines value as an estimate and confirming it algebraically. Counting is fixed by combining the direction with the vertex's height relative to the axis. Interpreting is fixed by writing down what the variable measures before solving anything. 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 one sentence saying why an x-intercept is a solution, and beside it draw a parabola crossing the axis twice with both intercepts circled and labelled as roots. Underneath, take the equation x squared minus x equals two through the whole method: rearrange it, name the related function, find its vertex, build a small table, sketch it, read the intercepts, and check both by substituting into the equation as first written with the brackets shown. Beneath that, draw the three cases for the number of crossings side by side and write next to each where the vertex sits relative to the axis. In the lower half, sketch a bridge cable model with x measured from the middle, mark both intercepts, and write out the doubling step that turns them into a span. Finally, in the margin, write the one-line rule for which expression gets graphed.
Your check should use the original equation on both substitutions. If you find yourself checking against the rearranged version, you are testing the reading only and not the rearrangement that most often goes wrong.
Recap
Five things, and the first is the idea the whole lesson rests on.
| If the question says | Your first move is |
|---|---|
| Solve by graphing | Rearrange so one side is nought |
| Estimate the solutions | Read the x-intercepts |
| Check your solutions | Substitute into the original equation |
| How many real solutions? | Count crossings, or locate the vertex |
| A model asks for a distance | Interpret both intercepts before answering |
Lesson 9.6 removes the estimating. The quadratic formula gives the exact roots of every quadratic equation in one step, and the picture built here is what makes its plus-or-minus sign and its two answers make sense.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.5 Solving Quadratic Equations by Graphing §9.5, pp. 526-532 — everything on these slides traces back here
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