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2.10 — I've Got a Fill-in
A quadratic function can be described by several equivalent equations, a table, or a graph. Each representation emphasizes different information. The most useful first step is to decide what the problem is asking you to find.
1. In class: A Practice Understanding Task
Three forms reveal different features
Consider the same quadratic written three ways:
\[f(x)=x^2+2x-15=(x+5)(x-3)=(x+1)^2-16.\]
Expanding either of the last two expressions produces the first. They are equivalent functions: they give the same output for every real input.
| Form | What it reveals quickly | Example |
|---|---|---|
| Standard \(ax^2+bx+c\) | \(y\)-intercept \((0,c)\); convenient for evaluating at a given input | \(x^2+2x-15\): \((0,-15)\) |
| Factored \(a(x-r)(x-s)\) | Real zeros and x-intercepts when real factors exist | \((x+5)(x-3)\): x-intercepts \((-5,0),(3,0)\) |
| Vertex \(a(x-h)^2+k\) | Vertex \((h,k)\), axis \(x=h\), maximum/minimum value | \((x+1)^2-16\): vertex \((-1,-16)\) |
The coefficient \(a\) is the same in every form; its sign gives the opening direction. A parabola opening upward has a minimum, and one opening downward has a maximum. A quadratic may have zero, one, or two real x-intercepts; in particular, not every quadratic has a factored form using real linear factors.
Given \(f(x)=2(x-4)^2-18\), find the vertex immediately: \((4,-18)\). The axis is \(x=4\). To find zeros, solve \(2(x-4)^2-18=0\): \((x-4)^2=9\), so \(x=1,7\). Thus factored form is \(2(x-1)(x-7)\). To find the y-intercept, evaluate at zero: \(f(0)=2(16)-18=14\). The expanded standard form is \(2x^2-16x+14\).
Move between forms without changing the function
A. Factored to standard: multiply
\[y=-2(x-3)(x+1).\]
Multiply the binomials first: \((x-3)(x+1)=x^2-2x-3\). Then distribute \(-2\): \[\boxed{y=-2x^2+4x+6}.\]
Quick check: both expressions give \(y=6\) at \(x=0\).
B. Standard to factored: reverse multiplication
Factor \(y=x^2-2x-24\). Find two integers whose product is \(-24\) and whose sum is \(-2\): \(-6\) and \(4\). Hence \(\boxed{y=(x-6)(x+4)}\), giving zeros \(6\) and \(-4\).
C. Standard to vertex: complete the square
Start with \(y=2x^2+12x+13\). Factor 2 from the x-terms: \(2(x^2+6x)+13\). Half of 6 is 3, so add and subtract 9 inside the parentheses:
\[\begin{aligned}y&=2\bigl[(x+3)^2-9\bigr]+13\\&=2(x+3)^2-18+13\\&=\boxed{2(x+3)^2-5}.\end{aligned}\]
The vertex is \((-3,-5)\). Notice that the 9 added inside parentheses contributes \(2\cdot9=18\), so we must subtract 18—not 9—from the outside result.
D. Vertex to standard and, when possible, to factored
\[y=-(x-5)^2=-[x^2-10x+25]=\boxed{-x^2+10x-25}.\]
Factored form is \(\boxed{-(x-5)(x-5)}\): a repeated root at \(x=5\). The parabola touches, rather than crosses, the x-axis at its vertex.
Work with graphs and tables
A quadratic's vertex is its turning point. In vertex form \(a(x-h)^2+k\), plot the vertex first. Then choose inputs equally spaced on either side of \(h\). Symmetric x-values give identical y-values. For example, with \(y=-3(x-1)^2+3\):
| \(x\) | −1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|
| \(y\) | −9 | 0 | 3 | 0 | −9 |
| First differences | \(+9,\ +3,\ -3,\ -9\) | ||||
| Second differences | \(-6,\ -6,\ -6\) | ||||
A first difference is the change between two consecutive outputs. A second difference is the difference between consecutive first differences. For a quadratic \(y=ax^2+bx+c\) sampled at equally spaced x-values of step \(d\), the second differences equal \(2ad^2\). In particular, when \(d=1\), they equal \(2a\).
Suppose the outputs at x = 0, 1, 2, 3, 4 are \(7,2,-1,-2,-1\). First differences are \(-5,-3,-1,1\). Second differences are \(2,2,2\), so if these data are generated by a quadratic and the input step is 1, \(a=1\). The smallest table value is \(-2\) at \(x=3\), with symmetric values at x=2 and x=4. The vertex form is \(\boxed{y=(x-3)^2-2}\). Expansion gives \(y=x^2-6x+7\). Check x=0: \(9-2=7\).
When the worksheet asks you to recover a quadratic from a graph or table, assume the representation is intended to describe a quadratic function. Read the grid scale carefully when estimating coordinates. Exact equations require exact graph information; a small printed graph may make some measurements approximate. A finite set of table entries, even with constant second differences, does not prove the unknown function is quadratic for all x.
2. Ready — Which form best answers a real-world question?
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
Suppose a ball's height in meters after t seconds is given equivalently by
\[H(t)=-4.9t^2+19.6t+58.8\] \[=-4.9(t-6)(t+2)\] \[=-4.9(t-2)^2+78.4.\]
Use the factored form to find when the ball reaches the water: the physically meaningful time is \(t=6\) seconds, not \(t=-2\), because the experiment starts at \(t=0\). Use the vertex form to read its maximum height, \(78.4\) meters, attained at \(t=2\) seconds. Use the standard form to see the cliff height at \(t=0\), which is \(58.8\) meters. To find height at an arbitrary specified time, any form works by substitution; standard form is often the most straightforward numerically.
At \(t=3.5\) seconds, vertex form gives \[H(3.5)=-4.9(3.5-2)^2+78.4=-4.9(2.25)+78.4=\boxed{67.375\text{ m}}.\] This substitution is equally valid in the other forms. The height is above the lake, not above the cliff.
Check your work — Ready 1–4 answer key
- Factored form \(-4.9(t-6)(t+2)\) exposes the zeros. The ball reaches the water after 6 seconds; −2 is before impact.
- Vertex form \(-4.9(t-2)^2+78.4\) shows a maximum of 78.4 meters at 2 seconds, since the squared term has a negative coefficient.
- Any form works; vertex form is convenient because \(3.5-2=1.5\). Height \(78.4-4.9(1.5)^2=67.375\) meters.
- Standard form shows the initial height as its constant term: 58.8 meters when \(t=0\).
3. Set — Assemble the full set of representations
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
For the shared ideas used in these exercises, revisit the In class explanation.
Given \(y=(x+4)(x-2)\), expand to standard: \(y=x^2+2x-8\). Complete the square: \(y=(x+1)^2-9\). The vertex is \((-1,-9)\); the x-intercepts are \((-4,0)\) and \((2,0)\). Use x-values \(-3,-2,-1,0,1\) around the vertex to get outputs \(-5,-8,-9,-8,-5\). First differences \(-3,-1,1,3\); second differences \(2,2,2\). A graph should be symmetric around \(x=-1\).
Given \(y=-2(x-3)^2+8\), expand: \(y=-2x^2+12x-10\). To factor, set up a difference of squares: \(-2[(x-3)^2-4]=-2(x-5)(x-1)\). Thus the zeros are 1 and 5. The vertex is \((3,8)\), and sample x-values 1,2,3,4,5 yield outputs \(0,6,8,6,0\). First differences \(6,2,-2,-6\); second differences \(-4,-4,-4\).
Take \(y=x^2+2x+4=(x+1)^2+3\). The vertex is \((-1,3)\), above the x-axis, and the parabola opens upward. There are no real x-intercepts, so there is no factorization into real linear factors. Still, it has a perfectly valid standard form, vertex form, graph, and table.
Check your work — Set 5–9 answer key
- (a) \(y=x^{2} + 2 x - 15\). (b) \(y=\left(x + 1\right)^{2} - 16\). (c) \(y=\left(x - 3\right) \left(x + 5\right)\). (d) Points \(\left( -3, \ -12\right),\ \left( -2, \ -15\right),\ \left( -1, \ -16\right),\ \left( 0, \ -15\right),\ \left( 1, \ -12\right)\); first differences \(-3,\ -1,\ 1,\ 3\); second differences all \(2\). (e) Plot not shown; vertex \(\left( -1, \ -16\right)\).
- (a) \(y=- 3 x^{2} + 6 x\). (b) \(y=3 - 3 \left(x - 1\right)^{2}\). (c) \(y=- 3 x \left(x - 2\right)\). (d) Points \(\left( -1, \ -9\right),\ \left( 0, \ 0\right),\ \left( 1, \ 3\right),\ \left( 2, \ 0\right),\ \left( 3, \ -9\right)\); first differences \(9,\ 3,\ -3,\ -9\); second differences all \(-6\). (e) Plot not shown; vertex \(\left( 1, \ 3\right)\).
- (a) \(y=- x^{2} + 10 x - 25\). (b) \(y=- \left(x - 5\right)^{2}\). (c) \(y=- \left(x - 5\right)^{2}\). (d) Points \(\left( 3, \ -4\right),\ \left( 4, \ -1\right),\ \left( 5, \ 0\right),\ \left( 6, \ -1\right),\ \left( 7, \ -4\right)\); first differences \(3,\ 1,\ -1,\ -3\); second differences all \(-2\). (e) Plot not shown; vertex \(\left( 5, \ 0\right)\).
- (a–d) The supplied graph has no labeled coordinate scale, so exact equations and a numerical table cannot be determined uniquely. If each grid interval means one unit on both axes, the intended forms appear to be \(y=x^2+5x+6=(x+\frac{5}{2})^2-\frac{1}{4}=(x+3)(x+2)\); vertex \((-\frac{5}{2},-\frac{1}{4})\); points \((-\frac{9}{2},\frac{15}{4}),(-\frac{7}{2},\frac{3}{4}),(-\frac{5}{2},-\frac{1}{4}),(-\frac{3}{2},\frac{3}{4}),(-\frac{1}{2},\frac{15}{4})\); first differences \(-3,-1,1,3\); second differences 2. These values depend on that unlabeled-scale assumption. (e) Graph supplied in the packet.
- (a) \(y=2x^2-12x+12\). (b) \(y=2(x-3)^2-6\). (c) Skip as directed. (d) First differences \(-10,-6,-2,2,6,10\); second differences all 4; vertex \((3,-6)\). (e) Plot not shown.
4. Go review — Factoring and verifying
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
When you verify a factorization, multiply the proposed factors and combine like terms. For monic quadratics \(x^2+bx+c\), integer factoring seeks integers \(p,q\) with \(p+q=b\) and \(pq=c\). Special patterns include \(u^2-v^2=(u-v)(u+v)\) and \(u^2\pm2uv+v^2=(u\pm v)^2\).
Verify \(x^2+9x+14=(x+7)(x+2)\). Multiply: \[(x+7)(x+2)=x^2+2x+7x+14=x^2+9x+14.\] Every coefficient matches, so the factorization is correct.
Factor \(z^2-4z-45\). Look for integers that multiply to −45 and add to −4. They are −9 and 5, so \(\boxed{(z-9)(z+5)}\). Verify: \(z^2+5z-9z-45=z^2-4z-45\).
Factor \(16p^2-81\). This is the difference of squares \((4p)^2-9^2\), so \(\boxed{(4p-9)(4p+9)}\). The cross terms cancel when multiplied.
Investigate \(q^2+5q+1\). The only integer factor pair of 1 is 1 and 1 (or −1 and −1); neither gives middle coefficient 5. Thus the expression is not factorable using integer-coefficient linear factors. It does have real roots \(\frac{-5\pm\sqrt{21}}2\), so “not factorable” must always specify the intended number system.
The expression \(x^2+10x+25=(x+5)^2\) represents the area of a square of side \(x+5\), provided the side length is nonnegative. Compare \(x^2+10x+24=(x+4)(x+6)\), generally a rectangle rather than a square. A symbolic rectangle description also requires dimensions that are nonnegative for the x-values in the physical context.
For problems 14–22, interpret “factor, if possible” as factor over the integers unless instructed otherwise. For problem 24, expressions are areas rather than literal side lengths; factors can represent rectangle side lengths when both are nonnegative on the chosen domain. Failing to factor over the integers does not prove that a real rectangle model is impossible.
Check your work — Go 10–24 answer key
- \((x+16)(x-4)=x^2-4x+16x-64=x^2+12x-64\).
- \((x+8)(x-8)=x^2-8x+8x-64=x^2-64\).
- \((x+16)(x+4)=x^2+4x+16x+64=x^2+20x+64\).
- \((x-8)(x-8)=x^2-8x-8x+64=x^2-16x+64\).
- \((x-2)(x-3)\).
- \((x-1)(x-6)\).
- \((x-9)(x+4)\).
- \((m+7)(m+9)\).
- Not factorable over the integers: its discriminant is 13, not a square.
- Not factorable over the integers: its discriminant is 41, not a square.
- \((x+7)^2\).
- \((x-3)(x+3)\).
- Not factorable over the integers: its discriminant is 109, not a square.
- #20 is \((x+7)^2\), a square with side \(|x+7|\). If “above” includes the verification exercises, #13 is also \((x-8)^2\).
- The expressions are areas, not side lengths. Under integer linear factoring, #18, #19, and #22 do not supply such side factors. They can still be areas of rectangles with real side lengths wherever the area is positive (for instance width 1 and length equal to the area). Factors from the other problems give physical side lengths only on domains where the chosen lengths are positive.
Additional Learning Resources
These links offer instruction or practice beyond the packet. Use the indicated relevant portions; Professor Leonard's lecture is optional and longer than this day's work.
| Resource and direct link | What it teaches | Scope / What to skip |
|---|---|---|
| Paul's Online Math Notes — Algebra §4.2 Parabolas | Graph features, vertex form, standard form, intercepts, examples | Close match to main and Set; emphasize graphing, intercepts, and rewriting, not unrelated conics. |
| Paul's Online Math Notes — §4.2 Practice Problems (solutions) | Problems 1–7 graphing; 8–10 converting to vertex form, with solutions | Close match; these are especially useful for independent practice. |
| OpenStax Algebra 1 — 7.16.1 Three Forms of Quadratic Expressions | Choosing a form to locate vertex and intercepts | Close match to Ready and the overall lesson. |
| Professor Leonard — Graphing Quadratic Functions (Precalculus / College Algebra 24) | Long-form classroom explanation of graphing using vertex form, completing the square, intercepts | Significantly broader. Optional extended learning: watch the sections on vertex, graphing and intercepts. The lecture also uses the quadratic formula and additional algebra; those portions are optional. No precise timestamps were verified. |
Teacher's review — Material issues only
Set 8 and Set 9: The representations are provided without an explicit assumption that the relationship is quadratic. The tasks intend a quadratic model; state this assumption. An unlabeled or tightly scaled graph may also limit the precision with which a coefficient can be recovered.
Go 14–24: “Some will not factor” does not specify the coefficient number system. Several examples are irreducible over the integers but factor over the reals. Problem 24 also conflates a quadratic area expression with possible rectangle side lengths; require nonnegative lengths on a stated input domain.
Curriculum coverage — Section 2.10 only
| Packet exercises | Skill | Where addressed |
|---|---|---|
| Main task 1–7 (pp. 58–61) | Recover forms from standard equations and graph; intercepts and vertices | In class, examples 1–6 |
| Ready 1–4 (p. 62) | Choose a useful form for flight-time, maximum, specified height, cliff height | Section 2, example 7 |
| Set 5–9 (pp. 62–64) | Standard, vertex, factored, tables and graphs, first/second differences | In class and Section 3, examples 6 and 8–10, clarification |
| Go 10–13 (p. 64) | Verify factorizations by multiplying | Section 4, Problem A |
| Go 14–22 (p. 64) | Integer factoring, difference of squares, recognizing non-integer-factorable trinomials | Section 4, Problems B–D |
| Go 23–24 (p. 64) | Perfect-square area and rectangle dimensions, physical restrictions | Section 4, Problem E and clarification |
Curriculum source: Mathematics Vision Project, Secondary Math II, Module 2, §2.10, packet pp. 58–64. Graphs are embedded SVGs with clipping. Mathematical notation loads through MathJax; an internet connection is required.
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