Building a quadratic function from a vertex and a point, from two x-intercepts and a point, and from three arbitrary points using a system of equations; choosing the right form for the information given; and finding a best-fitting quadratic model from data.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 4 — Quadratic Functions and Factoring
Write Quadratic Functions and Models
Objectives
Five outcomes. The whole chapter has been reading parabolas; this lesson writes them.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.10 Write Quadratic Functions and Models §4.10, pp. 309-313 — the lesson these objectives are drawn from
Warm-up
Every lesson so far has started with a function and asked about its graph. This one runs the other way.
Discussion prompt
Lesson 4.2 read the vertex off y equals a times the quantity x minus h squared, plus k. If you are told the vertex is (1, negative 2), how much of that function do you already know, and what is still missing?
Hint: Count the unknowns.
Answer:
\[ y = a(x - 1)^2 - 2 \]
Two of the three unknowns are settled: h is 1 and k is negative 2. Only a is left, and one more point on the parabola is enough to find it. That is the whole shape of this lesson — pick the form that already contains what you were given, and solve for what is left.
Concept
A quadratic has three unknowns whichever form it is written in, so three pieces of information determine it. What changes is how much work is left: a form that already displays the given features leaves only the leading coefficient to find, while three arbitrary points require a full system of equations.
best-fitting quadratic model — The quadratic function produced by quadratic regression on a set of data. It generally passes through none of the data points, but comes as close to all of them as possible.
\[ a(x-h)^2 + k, \quad a(x-p)(x-q), \quad ax^2 + bx + c \]
The three forms are the ones from Lesson 4.2, now used as templates to be filled in rather than as arrangements to be read.
Figure (svg): A decision table matching what is given about a parabola with the form to start from
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.10 Write Quadratic Functions and Models §4.10, pp. 309-310
Section
Section 1
Concept
If the vertex is given, substitute it for h and k in vertex form. That leaves a single unknown, the leading coefficient, which one more point on the parabola determines.
\[ y = a(x - h)^2 + k \]
The extra point must not be the vertex itself: substituting the vertex makes the bracket zero and the equation reduces to k equals k, which says nothing about a.
Figure (svg): A parabola with its vertex and one other point marked, used to determine the leading coefficient
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.10 Write Quadratic Functions and Models §4.10, pp. 309-309 — Write a quadratic function in vertex form
Picture it
Example 1: vertex (1, negative 2), passing through (3, 2).
Figure (svg): A parabola with its vertex and one other point marked, used to determine the leading coefficient
Substituting the vertex fills in h and k; substituting the extra point gives a single linear equation in a, which solves to 1.
Worked example
Example 1. Two substitutions, one solve.
\[ \text{Write a quadratic with vertex } (1, -2) \text{ through } (3, 2). \]
Choose the form
Why: The vertex is given, so vertex form already contains it.
\[ y = a(x - h) ^{2} + k \]
Substitute the vertex
Why: H is 1 and k is negative 2.
\[ y = a(x - 1) ^{2} - 2 \]
Substitute the other point
Why: X is 3 and y is 2.
\[ 2 = a(3 - 1) ^{2} - 2 \]
Solve for a
Why: Three minus 1 is 2, squared is 4, so 2 equals 4a minus 2, giving 4a equals 4.
\[ a = 1 \]
Figure (svg): The solution to Worked example write in vertex form shown as a ladder of expressions, one row per algebraic move
\[ y = (x - 1)^2 - 2 \]
Verify: substitute both given facts back
Why: At x equal to 1 the function gives negative 2, so the vertex is right. At x equal to 3 it gives 4 minus 2, or 2, so the point is on the curve. Since a came out as 1, the parabola is the parent shape simply translated — which the picture confirms.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.10 Write Quadratic Functions and Models §4.10, pp. 309-309
Matching
Substitute the vertex, then the point.
Match the pairs
Why: Two of these have a equal to 1, so they are the parent parabola translated; one has a negative a and opens downward; and one has a vertex at the origin, so both h and k are zero and the form collapses to a times x squared.
Worked example
Guided Practice 1 and 2. One of them opens downward.
\[ \text{Vertex } (4, -5) \text{ through } (2, -1); \quad \text{vertex } (-3, 1) \text{ through } (0, -8). \]
First: substitute the vertex
Why: The form becomes a times the quantity x minus 4, squared, minus 5.
\[ y = a(x - 4) ^{2} - 5 \]
First: use the point
Why: At x equal to 2 the bracket is negative 2, whose square is 4, so negative 1 equals 4a minus 5.
\[ 4 a = 4, a = 1 \]
Second: substitute the vertex
Why: The form becomes a times the quantity x plus 3, squared, plus 1.
\[ y = a(x + 3) ^{2} + 1 \]
Second: use the point
Why: At x equal to 0 the bracket is 3, whose square is 9, so negative 8 equals 9a plus 1.
\[ 9 a = -9, a = -1 \]
Figure (svg): The solution to Worked example two more vertices shown as a ladder of expressions, one row per algebraic move
\[ y = (x-4)^2 - 5, \qquad y = -(x+3)^2 + 1 \]
Verify: check the direction of opening
Why: The second has a equal to negative 1, so it opens downward and its vertex at (negative 3, 1) is a maximum. That is consistent with the given point (0, negative 8) lying well below the vertex, which would be impossible for an upward parabola. The sign of a is always worth sanity-checking against the picture.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.10 Write Quadratic Functions and Models §4.10, pp. 310-310
Trap
\[ y = a(x-1)^2 - 2, \text{ vertex } (1,-2) \]
Substitute the vertex again to find a
Why: The vertex is a point on the parabola, so it is used as the second substitution.
\[ -2 = a(1-1)^2 - 2 \;\Longrightarrow\; -2 = -2 \quad \text{(no information)} \]
The equation is true for every value of a, so nothing has been learned. Any parabola with that vertex satisfies it.
\[ 2 = a(3-1)^2 - 2 \;\Longrightarrow\; a = 1 \]
Use a point that is not the vertex
Why: The bracket must be non-zero for a to appear in the equation at all.
\[ y = (x-1)^2 - 2 \]
The general rule is that each new piece of information must be independent of the ones already used. A vertex fixes two numbers; a second point fixes the third; a third point would then be redundant or contradictory.
Fill the middle
Example 1, at the third step.
Fill in the blanks
2 = a(3 - 1)^2 - 2 \;\Longrightarrow\; 2 = 4a - 2 \;\Longrightarrow\; a = 1
Why: Three minus 1 is 2, and 2 squared is 4, so the equation reads 2 equals 4a minus 2. Adding 2 gives 4a equals 4 and a equals 1. Note that the bracket is evaluated and squared before a multiplies it, which is the order of operations rather than a special rule.
Ranking
Writing a quadratic from a vertex and a point.
Put in order
Why: The vertex goes in before the point, because until h and k are filled the equation from the point would have three unknowns instead of one. The check at the end is quick and worth doing: substituting the vertex should return k exactly.
Prediction
Commit before reasoning.
Predict first
You are told only that a parabola has vertex (1, negative 2). How many such parabolas are there?
Correct: Infinitely many, one for each value of a.
\[ y = a(x-1)^2 - 2 \text{ for any } a \neq 0 \]
Why: The vertex fixes h and k but says nothing about the width or direction, so every non-zero a gives a different parabola with that same vertex. One more point picks out exactly one of them. This is the same counting argument that runs through the whole lesson: three unknowns need three independent pieces of information, and a vertex is worth two of them.
Section
Section 2
Concept
If both x-intercepts are given, substitute them for p and q in intercept form. As with vertex form, that leaves only the leading coefficient, which one more point determines.
\[ y = a(x - p)(x - q) \]
Substitute carefully: an intercept at negative 1 goes into the form as x minus negative 1, that is x plus 1. The book flags this in an Avoid Errors note.
Figure (svg): A downward parabola through two marked x-intercepts and one other point
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.10 Write Quadratic Functions and Models §4.10, pp. 309-309 — Write a quadratic function in intercept form
Picture it
Example 2: intercepts at negative 1 and 4, passing through (3, 2).
Figure (svg): A downward parabola through two marked x-intercepts and one other point
The leading coefficient comes out as negative one half, so the parabola opens downward and is wider than the parent — both visible in the picture.
Worked example
Example 2. Watch the sign of the negative intercept.
\[ \text{Write a quadratic with x-intercepts } -1 \text{ and } 4 \text{ through } (3, 2). \]
Choose the form
Why: Both x-intercepts are given, so intercept form already contains them.
\[ y = a(x - p) (x - q) \]
Substitute the intercepts
Why: P is negative 1, so the first bracket is x plus 1; q is 4, so the second is x minus 4.
\[ y = a(x + 1) (x - 4) \]
Substitute the other point
Why: X is 3 and y is 2.
\[ 2 = a(3 + 1) (3 - 4) \]
Solve for a
Why: Four times negative 1 is negative 4, so 2 equals negative 4a.
\[ a = -\frac{1}{2} \]
Figure (svg): The solution to Worked example write in intercept form shown as a ladder of expressions, one row per algebraic move
\[ y = -\tfrac{1}{2}(x + 1)(x - 4) \]
Verify: check all three given facts
Why: At x equal to negative 1 and at x equal to 4 the product vanishes, so both intercepts are right. At x equal to 3 the value is negative one half times 4 times negative 1, which is 2. All three check, and the negative a matches a parabola whose vertex, at x equal to 1.5, is above the axis.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.10 Write Quadratic Functions and Models §4.10, pp. 309-309
Fill the middle
Example 2, at the second step.
Fill in the blanks
\text+ -1 \text___ 4 \;\Longrightarrow\; y = a(x ___ 1)(x - 4)
Why: The bracket must vanish at x equal to negative 1, and x plus 1 does exactly that. The form is written with a minus sign, so substituting p equal to negative 1 gives x minus negative 1, which is x plus 1. Asking what makes the bracket zero avoids the double-negative bookkeeping entirely.
Worked example
Guided Practice 3.
\[ \text{Write a quadratic with x-intercepts } -2 \text{ and } 5 \text{ through } (6, 2). \]
Substitute the intercepts
Why: The brackets are x plus 2 and x minus 5.
\[ y = a(x + 2) (x - 5) \]
Substitute the point
Why: At x equal to 6 the brackets are 8 and 1.
\[ 2 = a(8) (1) \]
Solve for a
Why: Two equals 8a, so a is one quarter.
\[ a = \frac{1}{4} \]
Write the function
Why: Positive a means the parabola opens upward and is wider than the parent.
\[ y = (\frac{1}{4}) (x + 2) (x - 5) \]
Figure (svg): The solution to Worked example another pair of intercepts shown as a ladder of expressions, one row per algebraic move
\[ y = \tfrac{1}{4}(x + 2)(x - 5) \]
Verify: find the vertex as a further check
Why: The axis of symmetry is the average of the intercepts, which is 1.5, and substituting gives one quarter times 3.5 times negative 3.5, or about negative 3.06. The vertex is below the axis, as an upward parabola with two real intercepts requires. That consistency is a good sign the coefficient is right.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.10 Write Quadratic Functions and Models §4.10, pp. 310-310
Error analysis
A student substitutes an intercept of negative 1 into intercept form.
Annotate
On: \( y = a(x - 1)(x - 4) \text{ for x-intercepts } -1 \text{ and } 4 \)
Same sign trap as reading h in Lesson 4.2. Ask what value makes each bracket zero, and the sign takes care of itself.
Sorting
Solve for a and read its sign.
Sort into buckets
Sort each set of conditions by the direction the parabola opens.
The sign of a is not a separate decision; it falls out of the arithmetic. But checking it against a rough sketch catches substitution errors quickly.
Comparison
Fill the blanks. Two forms, the same amount of work.
Comparison matrix
| Question | Vertex form | Intercept form |
|---|---|---|
| What you must be given | the vertex, plus a point | both x-intercepts, plus a point |
| Unknowns after substituting | one, namely a | one, namely a |
| Always available? | yes, every parabola has a vertex | no, some have no x-intercepts |
| Sign trap | h flips: x + 2 means h = -2 | p and q flip the same way |
The third row is the real difference: intercept form does not exist for a parabola that misses the axis, so it is available only when the problem hands you two crossings.
Prediction
Commit before reasoning.
Predict first
You are told a parabola crosses at x equal to 3 and passes through (5, 8). Can you write it?
Correct: No — that is only two pieces of information, and three are needed.
\[ y = a(x - 3)(x - q): \; \text{two unknowns, one equation} \]
Why: Intercept form needs both crossings, and a single intercept plus a point leaves two unknowns among a, and the missing q. Infinitely many parabolas pass through those two points. A third independent fact — the other intercept, the vertex, or a third point — is required, and that requirement is the same in every form.
Section
Section 3
Concept
When neither the vertex nor the intercepts are given, use standard form. Substituting each point produces one linear equation in a, b and c, and the resulting system of three equations in three unknowns is solved by the methods of Lesson 3.4.
\[ y = ax^2 + bx + c \]
A point whose x value is zero is worth substituting first: it gives c immediately, since both variable terms vanish, and reduces the system to two equations in two unknowns.
Figure (svg): Three plotted points and the single parabola passing through all of them
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.10 Write Quadratic Functions and Models §4.10, pp. 310-310 — Write a quadratic function in standard form
Picture it
Example 3: the points (negative 1, negative 3), (0, negative 4) and (2, 6).
Figure (svg): Three plotted points and the single parabola passing through all of them
The middle point has x equal to zero, so it gives c equal to negative 4 in one line. The other two then form a two-by-two system solved by elimination.
Worked example
Example 3, in the book's three steps.
\[ \text{Write a quadratic through } (-1,-3), \; (0,-4) \text{ and } (2,6). \]
Substitute each point into standard form
Why: The first gives negative 3 equals a minus b plus c; the second gives negative 4 equals c; the third gives 6 equals 4a plus 2b plus c.
Use the easy equation first
Why: The second says c is negative 4 outright, because both variable terms vanish at x equal to zero.
\[ c = -4 \]
Substitute c into the other two
Why: The first becomes a minus b equals 1; the third becomes 4a plus 2b equals 10.
\[ a - b = 1, 4 a + 2 b = 10 \]
Solve by elimination
Why: Doubling the first gives 2a minus 2b equals 2; adding to the third gives 6a equals 12, so a is 2 and then b is 1.
\[ a = 2, b = 1 \]
Figure (svg): The solution to Worked example write in standard form shown as a ladder of expressions, one row per algebraic move
\[ y = 2x^2 + x - 4 \]
Verify: substitute all three points
Why: At x equal to negative 1: 2 minus 1 minus 4 is negative 3. At x equal to 0: negative 4. At x equal to 2: 8 plus 2 minus 4 is 6. All three match, so the parabola really does pass through every given point — which is what distinguishes an exact fit from the regression of the last idea.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.10 Write Quadratic Functions and Models §4.10, pp. 310-310
Fill the middle
Example 3, at the second equation.
Fill in the blanks
-4 = a(0)^2 + b(0) + c \;\Longrightarrow\; c = -4
Why: At x equal to zero both the squared and the first-power terms are zero, so the equation reduces to y equals c. This is the y-intercept fact from Lesson 4.1 doing useful work: any point on the vertical axis hands you c for free, which is why it is worth substituting first.
Worked example
Guided Practice 4 and 6. Both have a point on the vertical axis.
\[ \text{Through } (-1,5), (0,-1), (2,11); \quad \text{through } (-1,0), (1,-2), (2,-15). \]
First: read c from the axis point
Why: At x equal to 0 the value is negative 1, so c is negative 1.
\[ c = -1 \]
First: form and solve the two-by-two
Why: The other points give a minus b equals 6 and 4a plus 2b equals 12, that is 2a plus b equals 6; adding gives 3a equals 12.
\[ a = 4, b = -2 \]
Second: no axis point, so subtract equations
Why: The points at x equal to negative 1 and 1 give a minus b plus c equals 0 and a plus b plus c equals negative 2; subtracting gives 2b equals negative 2.
\[ b = -1 \]
Second: finish the system
Why: Then a plus c equals negative 1, and the third point gives 4a plus c equals negative 13; subtracting gives 3a equals negative 12.
\[ a = -4, c = 3 \]
Figure (svg): The solution to Worked example two more systems shown as a ladder of expressions, one row per algebraic move
\[ y = 4x^2 - 2x - 1, \qquad y = -4x^2 - x + 3 \]
Verify: substitute one point into each
Why: For the first, at x equal to 2: 16 minus 4 minus 1 is 11, matching. For the second, at x equal to 2: negative 16 minus 2 plus 3 is negative 15, matching. Note the second problem had no point on the vertical axis, so c had to be found last rather than first — the method adapts without changing.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.10 Write Quadratic Functions and Models §4.10, pp. 310-310
Error analysis
A student is given three points but uses only two of them.
Annotate
On: \( (-1,-3) \text{ and } (0,-4) \;\Longrightarrow\; c = -4, \; a - b = 1 \;\Longrightarrow\; a = 2, b = 1 \)
Three unknowns need three independent equations. If you find yourself choosing a value rather than solving for it, a piece of information has gone unused.
Ranking
Writing a quadratic through three points.
Put in order
Why: Steps two and three are a shortcut that applies only when one of the points lies on the vertical axis; without one, the three-by-three system is solved directly by the methods of Lesson 3.4. The final check is not optional here — a system offers many places for an arithmetic slip, and substituting three points catches all of them.
Two truths and a lie
All three are about determining a parabola.
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. Knowing whether a is positive or negative is not the same as knowing its value, so two points plus a direction still leave a whole family of parabolas — every sufficiently steep upward one through those two points, for instance. Three genuine numerical facts are needed, and a direction is not one of them.
Prediction
Commit before reasoning.
Predict first
Could the points (1, 3), (1, 5) and (2, 4) lie on one parabola?
Correct: No — a function cannot send one input to two outputs.
\[ 3 = a + b + c \text{ and } 5 = a + b + c: \; \text{impossible} \]
Why: This is the vertical line test from Lesson 2.1. The input 1 appears twice with different outputs, so no function at all passes through all three points, quadratic or otherwise. Algebraically the system would produce two contradictory equations, and it is worth checking the x values before starting rather than discovering the contradiction three steps in.
Section
Section 4
Concept
All three forms describe the same kinds of function, so any of them could in principle be used. The one to choose is whichever already displays what you were given, because that leaves the fewest unknowns to solve for.
\[ \text{vertex} \to a(x-h)^2+k; \quad \text{intercepts} \to a(x-p)(x-q); \quad \text{points} \to ax^2+bx+c \]
The answer can always be converted afterwards: expanding vertex or intercept form gives standard form, and completing the square from Lesson 4.7 goes the other way.
Figure (svg): A decision table matching what is given about a parabola with the form to start from
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.10 Write Quadratic Functions and Models §4.10, pp. 309-310
Picture it
Four kinds of information, four starting points.
Figure (svg): A decision table matching what is given about a parabola with the form to start from
The first two rows leave a single unknown, the third leaves three, and the fourth hands the arithmetic to a calculator. Recognising which row you are in takes a second and saves several minutes.
Worked example
Example 2's parabola, rebuilt from three points instead.
\[ \text{Find the quadratic through } (-1,0), \; (4,0) \text{ and } (3,2) \text{ using standard form.} \]
Substitute all three points
Why: Zero equals a minus b plus c; zero equals 16a plus 4b plus c; 2 equals 9a plus 3b plus c.
Subtract the first from the second
Why: Fifteen a plus 5b equals 0, so 3a plus b equals 0 and b equals negative 3a.
\[ b = -3 a \]
Substitute into the first
Why: A plus 3a plus c equals 0, so c equals negative 4a.
\[ c = -4 a \]
Use the third point
Why: Nine a minus 9a minus 4a equals 2, so negative 4a equals 2 and a is negative one half.
\[ a = -\frac{1}{2} \]
Assemble the answer
Why: Then b is 1.5 and c is 2.
\[ y = -0.5 x ^{2} + 1.5 x + 2 \]
Figure (svg): The solution to Worked example the same parabola, two routes shown as a ladder of expressions, one row per algebraic move
\[ y = -\tfrac{1}{2}x^2 + \tfrac{3}{2}x + 2 \]
Verify: expand the intercept-form answer
Why: Example 2 gave negative one half times the quantity x plus 1, times the quantity x minus 4. Expanding: the brackets multiply to x squared minus 3x minus 4, and negative one half of that is negative one half x squared plus 1.5x plus 2 — the same function. Both routes work; the intercept-form route took four lines instead of ten.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.10 Write Quadratic Functions and Models §4.10, pp. 309-310
Matching
Ask which form already displays what you were told.
Match the pairs
Why: The first two leave a single unknown and take three or four lines; the third needs a full system; and the fourth cannot be solved exactly at all, because twenty points almost never lie on one parabola. Each row is a different amount of work for the same kind of answer.
Worked example
Guided Practice 5. No point on the vertical axis makes it longer, but the method is unchanged.
\[ \text{Write a quadratic through } (-2,-1), \; (0,3) \text{ and } (4,1). \]
Read c from the axis point
Why: At x equal to 0 the value is 3.
\[ c = 3 \]
Substitute into the other two
Why: Four a minus 2b plus 3 equals negative 1 gives 2a minus b equals negative 2; 16a plus 4b plus 3 equals 1 gives 8a plus 2b equals negative 1.
Solve by substitution
Why: From the first, b equals 2a plus 2; substituting gives 8a plus 4a plus 4 equals negative 1.
\[ 12 a = -5 \]
Finish
Why: A is negative five twelfths, and b is negative five sixths plus 2, which is seven sixths.
\[ a = -\frac{5}{12}, b = \frac{7}{6} \]
Figure (svg): The solution to Worked example a system with fractions shown as a ladder of expressions, one row per algebraic move
\[ y = -\tfrac{5}{12}x^2 + \tfrac{7}{6}x + 3 \]
Verify: substitute the last point
Why: At x equal to 4: negative five twelfths times 16 is negative twenty thirds; seven sixths times 4 is fourteen thirds; and negative twenty thirds plus fourteen thirds is negative two, plus 3 gives 1 — matching the given point. Fractional coefficients are entirely normal when the points are not chosen to be tidy, and are not a sign of an error.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.10 Write Quadratic Functions and Models §4.10, pp. 310-310
Trap
\[ \text{vertex } (1,-2), \text{ through } (3,2) \]
Set up a system in a, b and c
Why: Standard form is the most familiar, so it is used regardless of what was given.
\[ \text{but the vertex is not a substitutable point pair alone} \quad \text{(stuck)} \]
A vertex is two facts, not one, and expressing that in standard form needs the extra equation negative b over 2a equals 1 — more work, and easy to forget.
\[ y = a(x-1)^2 - 2 \;\Longrightarrow\; 2 = 4a - 2 \;\Longrightarrow\; a = 1 \]
Start from the form that already contains the given feature
Why: Vertex form absorbs the vertex in one substitution, leaving a single unknown.
\[ y = (x-1)^2 - 2 = x^2 - 2x - 1 \]
Converting to standard form afterwards takes one expansion, so nothing is lost by starting elsewhere. The right question is not which form do I like but which form is the information already in.
Sorting
Read what is given, not what looks familiar.
Sort into buckets
Sort each set of conditions.
Two of these disguise the information in different vocabulary. Maximum value at an x is a vertex, and zeros are x-intercepts.
Fill the middle
Example 2's answer, expanded.
Fill in the blanks
-\tfrac2___(x+1)(x-4) = -\tfrac______(x^2 - 3x - 4) = -\tfrac______x^2 + \tfrac______x + ___
Why: Negative one half times negative 4 is positive 2, which is the y-intercept — and substituting x equal to zero into the original product confirms it: negative one half times 1 times negative 4 is 2. Converting to standard form is worth doing when a question asks for the y-intercept, which neither of the other forms displays.
Prediction
Commit before reasoning.
Predict first
Example 2 solved by intercept form and by a three-point system. Should the answers agree?
Correct: Yes — three facts determine one parabola, however you find it.
\[ -\tfrac{1}{2}(x+1)(x-4) = -\tfrac{1}{2}x^2 + \tfrac{3}{2}x + 2 \]
Why: The two intercepts and the extra point are three independent facts, and exactly one quadratic satisfies them, so both routes must land on it. What differs is the amount of algebra and the form the answer arrives in — one as a product, one as a sum — and expanding turns the first into the second. Having two routes is useful precisely because each checks the other.
Section
Section 5
Concept
Real data rarely lie exactly on a curve. Quadratic regression finds the parabola that comes closest to all the points at once, called the best-fitting quadratic model, and it is used the same way any other model is.
\[ y = -0.261x^2 + 22.6x + 23.0 \]
The procedure is the same as the linear regression of Lesson 2.6, with a quadratic option chosen instead of a linear one: enter the data, make a scatter plot, check the trend looks parabolic, and run the regression.
Figure (svg): A scatter plot of pumpkin launch distances against angle with the best-fitting parabola drawn through it
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.10 Write Quadratic Functions and Models §4.10, pp. 311-311 — Quadratic regression
Picture it
Example 4: six launch angles and the distances thrown.
Figure (svg): A scatter plot of pumpkin launch distances against angle with the best-fitting parabola drawn through it
The curve passes near every point and through none of them. Its vertex, at about 43 degrees, is the angle that throws a pumpkin farthest — a fact none of the six measurements states directly.
Worked example
Example 4, in the book's four steps.
\[ \text{Fit a quadratic to distances } 372, 462, 509, 501, 437, 323 \text{ at angles } 20 \text{ to } 70. \]
Enter the data into two lists
Why: One list holds the angles and the other the distances.
Make a scatter plot and inspect it
Why: The points rise, level off and fall, which is a parabolic trend rather than a linear one.
Run quadratic regression
Why: The calculator returns the coefficients of the best-fitting quadratic.
\[ y = -0.261 x ^{2} + 22.6 x + 23.0 \]
Graph the model with the data
Why: The curve passes close to all six points, confirming the fit is reasonable.
Figure (svg): The solution to Worked example find the best-fitting model shown as a ladder of expressions, one row per algebraic move
\[ y = -0.261x^2 + 22.6x + 23.0 \]
Verify: evaluate the model at a data point
Why: At 40 degrees the model gives negative 0.261 times 1600, or negative 417.6, plus 22.6 times 40, or 904, plus 23, which is about 509 feet — very close to the measured 509. At 70 degrees it gives about 328 against a measured 323. Being a few feet out at each point is what best fit means; passing through them exactly is not.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.10 Write Quadratic Functions and Models §4.10, pp. 311-311
Sorting
Count the given points against the three unknowns.
Sort into buckets
Sort each situation.
The dividing line is whether the number of facts exceeds three, and whether the facts are exact or measured.
Worked example
Guided Practice 7. The question is not in the data.
\[ \text{At what angle does the pumpkin travel farthest?} \]
Recognise what is being asked
Why: The greatest distance is the maximum of the model, which is the y value at its vertex.
Use the vertex formula from Lesson 4.1
Why: The axis of symmetry is at negative b over 2a.
\[ x = -\frac{22.6}{2(-0.261)} \]
Evaluate
Why: Twenty-two point six divided by 0.522 is about 43.3.
\[ \text{about } 43 ^\circ \]
Interpret
Why: None of the six measurements was taken at 43 degrees; the model supplies the answer between them.
\[ 43 ^\circ \]
Figure (svg): The solution to Worked example use the model to answer a question shown as a ladder of expressions, one row per algebraic move
\[ x = -\tfrac{b}{2a} \approx 43 \text{ degrees} \]
Verify: check against the measured data
Why: The two largest measured distances were at 40 and 50 degrees, at 509 and 501 feet, so the peak lies between them and nearer to 40 — consistent with 43. Substituting 43.3 into the model gives about 512 feet, slightly above every measurement, which is exactly what a peak between data points should look like.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.10 Write Quadratic Functions and Models §4.10, pp. 311-311
Error analysis
A student checks a regression model against one of the data points.
Annotate
On: \( \text{at } 70^\circ \text{ the model gives } 328 \text{ but the data say } 323, \text{ so the model is wrong} \)
Exact fit and best fit are different jobs. With exactly three points they coincide; with more, insisting on exactness usually means no quadratic exists at all.
Fill the middle
Guided Practice 7.
Fill in the blanks
x = -\frac0.522___ = \frac______} \approx 43
Why: Twice 0.261 is 0.522, and the two negatives cancel to leave a positive quotient of about 43.3. This is the vertex formula from Lesson 4.1 applied to a model rather than to a textbook function, which is the whole point of building the model in the first place.
Ranking
Finding and using a best-fitting model.
Put in order
Why: Step two is the one people skip and should not: if the scatter plot shows a straight-line trend, a quadratic model will still be produced and will still be misleading. Judging the fit visually in step four is the same check applied afterwards.
Prediction
Commit before reasoning.
Predict first
The model was fitted to angles from 20 to 70 degrees. What does it give at 90, and should you trust it?
Correct: About 940 feet, but that value should not be trusted.
\[ y(90) = -0.261(8100) + 22.6(90) + 23 \approx -57 \text{ feet} \]
Why: Substituting 90 gives negative 0.261 times 8100, or about negative 2114, plus 2034 plus 23, which is about negative 57 feet — a negative distance, which is physically meaningless. Extrapolating a model beyond the range of the data it was fitted to is where models most often fail, and a pumpkin launched straight up would in fact travel no horizontal distance at all. The model is honest only between 20 and 70 degrees.
Comparison
Fill the blanks. Every lesson answered a different question about the same curve.
Comparison matrix
| Question | Method | Lesson |
|---|---|---|
| What does it look like? | graph from standard, vertex or intercept form | 4.1 and 4.2 |
| Where does it cross zero? | factor, take roots, complete the square, or use the formula | 4.3 to 4.8 |
| Where is it positive? | critical values and interval tests | 4.9 |
| What function is it? | fit a form to the given information | 4.10 |
This lesson is the inverse of the first two: those went from a function to a picture, and this one goes from a picture, or from data, back to a function.
Pattern
One routine, branching on what you were given.
With more data points than unknowns, no exact fit exists: use regression, and never extrapolate the model beyond the range of the data it was built from.
OpenStax Algebra and Trigonometry 2e, §5.1 Quadratic Functions §5.1
Check
Vertex form. Two substitutions.
Check your understanding
Write a quadratic with vertex (1, -2) passing through (3, 2).
Answer: A
Why: Substituting the vertex gives y = a(x - 1)^2 - 2, and the point (3, 2) gives 2 = 4a - 2, so a = 1.
Check
Intercept form. Watch the negative intercept.
Check your understanding
Write a quadratic with x-intercepts -1 and 4 passing through (3, 2).
Answer: A
Why: The brackets are (x + 1) and (x - 4), and 2 = a(4)(-1) gives a = -1/2.
Check
Three points, one system.
Check your understanding
Write a quadratic in standard form through (-1, -3), (0, -4) and (2, 6).
Answer: A
Why: The point on the vertical axis gives c = -4, and the remaining system gives a = 2 and b = 1.
Real world
An architect is designing a parabolic arch for a doorway. The opening must be 8 feet wide at floor level and 10 feet high at its highest point, which is in the middle.
Discussion prompt
Write the arch's equation, taking the floor as the horizontal axis and the centre of the doorway as the origin. Then find how wide the arch is at a height of 6 feet.
Hint: With the origin at the centre, the two ground-level points are symmetric about it.
Answer:
\[ \text{intercepts } -4 \text{ and } 4, \text{ vertex } (0, 10) \;\Longrightarrow\; y = a(x+4)(x-4) \]
\[ 10 = a(4)(-4) = -16a \;\Longrightarrow\; a = -\tfrac{5}{8} \;\Longrightarrow\; y = -\tfrac{5}{8}(x^2 - 16) \]
\[ 6 = -\tfrac{5}{8}x^2 + 10 \;\Longrightarrow\; x^2 = \tfrac{32}{5} \;\Longrightarrow\; x \approx \pm 2.53 \]
The arch is about 5.06 feet wide at a height of 6 feet.
Two things are worth noticing. The vertex is on the vertical axis, so h is zero and either form would have worked equally well — the intercept route was chosen because the two ground points were given directly. And the last step is Lesson 4.5's square-root method, which is available precisely because the arch was placed symmetrically about the origin, killing the first-power term.
Commit first
Answer, then rate your confidence honestly.
Predict first
How many quadratic functions pass through the three points (0, 0), (1, 1) and (2, 4)?
Correct: Exactly one.
\[ (0,0), (1,1), (2,4) \;\Longrightarrow\; c = 0, \; a + b = 1, \; 4a + 2b = 4 \;\Longrightarrow\; a = 1, b = 0 \]
Why: Three points with distinct x values determine exactly one quadratic, and here it is y equals x squared. The points are not collinear — the differences 1 and 3 are not equal — so the leading coefficient does not come out as zero. If they had been collinear, the system would still have exactly one solution, but with a equal to zero, giving a line rather than a genuine parabola. Either way the count of solutions is one.
Explain it
They can graph a quadratic and are confused about being asked to produce one.
Discussion prompt
In four sentences or fewer, explain how you decide which form to start from, and why a vertex counts as two pieces of information.
Hint: Count unknowns against facts.
Answer:
Every quadratic has three unknown numbers in it, so you need three independent facts to pin it down. Start from whichever form already shows what you were told: vertex form if you were given the vertex, intercept form if you were given both crossings, and standard form otherwise.
A vertex counts as two facts because it supplies both h and k at once, which is why one extra point finishes the job. Two intercepts do the same thing, supplying p and q, so again a single extra point is enough.
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 choosing, write down what you were given in words before touching any algebra. For signs, ask what value makes each bracket zero rather than copying the number. For the system, look first for a point with x equal to zero, which gives c free. For regression, remember that the model passes near the data rather than through it, and never trust it outside the range it was fitted on. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Choose one parabola of your own with a positive leading coefficient and two x-intercepts, and hide it from yourself at the bottom of the page. Then, at the top, write down three different sets of information about it: its vertex plus one point, its two intercepts plus one point, and three ordinary points. In the middle of the page, reconstruct the function three times, once from each set, showing the substitutions and the algebra for each — including the full system for the third. Down the right side, expand all three answers into standard form and check that they are identical. At the bottom, uncover your original function and compare. In a margin, write one sentence saying which of the three routes took the least work and why.
If your three standard forms disagree, check the system first: it is the route with the most arithmetic and by far the most likely to contain the slip.
Recap
Five things, and together they close the chapter.
| If you are given | Then |
|---|---|
| A vertex and one point | Vertex form; solve for a |
| Two intercepts and one point | Intercept form; solve for a |
| Three ordinary points | Standard form; solve a system |
| A point with x = 0 | It gives c immediately |
| More than three data points | Use regression, not algebra |
| A question about a maximum | Find the model's vertex |
Chapter 4 began by graphing quadratics and ends by producing them from evidence. Chapter 5 raises the degree, and every technique here — factoring, the zero product property, systems of equations — reappears for polynomials of any degree.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.10 Write Quadratic Functions and Models §4.10, pp. 309-313 — everything on these slides traces back here
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